Intel Code Challenge Elimination Round (Div.1 + Div.2, combined)

A. Broken Clock

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.

You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format.

For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39.

Input

The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively.

The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes.

Output

The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them.

Examples

input

24
17:30

output

17:30

input

12
17:30

output

07:30

input

24
99:99

output

09:09

这么一道水题，我写的什么烂代码。。。

 #include<iostream>
 using namespace std;

 int n;
 int a,b;

 int main()
 {
     cin>>n;
     scanf("%d:%d",&a,&b);
     if(n==)
     {
         if(a==)cout<<"10:";
         else
         if(<=a&&a<)
             cout<<""<<a%<<':';
         else
         if(a>)
         {
             if(a%==)
                 cout<<"10:";
             else cout<<''<<a%<<':';
         }
         else cout<<a<<':';
         if(<=b&&b<) cout<<""<<b%;
         else
         if(b>=)
             cout<<''<<b%;
         else cout<<b;
     }
     if(n==)
     {
         if(<=a&&a<)
             cout<<""<<a%<<':';
         else
         if(a>=)
         {
             if(a%==)
                 cout<<"10:";
             else cout<<''<<a%<<':';
         }
         else cout<<a<<':';
         if(<=b&&b<) cout<<""<<b%;
         else
         if(b>=)
             cout<<''<<b%;
         else cout<<b;
     }
     return ;
 }

B. Verse Pattern

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a text consisting of n lines. Each line contains some space-separated words, consisting of lowercase English letters.

We define a syllable as a string that contains exactly one vowel any arbitrary number (possibly none) of consonants. In English alphabet following letters are considered to be vowels: 'a', 'e', 'i', 'o', 'u' and 'y'.

Each word of the text that contains at least one vowel can be divided into syllables. Each character should be a part of exactly one syllable. For example, the word "mamma" can be divided into syllables as "ma" and "mma", "mam" and "ma", and "mamm" and "a". Words that consist of only consonants should be ignored.

The verse patterns for the given text is a sequence of n integers p1, p2, ..., pn. Text matches the given verse pattern if for each i from 1 to n one can divide words of the i-th line in syllables in such a way that the total number of syllables is equal to pi.

You are given the text and the verse pattern. Check, if the given text matches the given verse pattern.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100) — the number of lines in the text.

The second line contains integers p1, ..., pn (0 ≤ pi ≤ 100) — the verse pattern.

Next n lines contain the text itself. Text consists of lowercase English letters and spaces. It's guaranteed that all lines are non-empty, each line starts and ends with a letter and words are separated by exactly one space. The length of each line doesn't exceed 100 characters.

Output

If the given text matches the given verse pattern, then print "YES" (without quotes) in the only line of the output. Otherwise, print "NO" (without quotes).

Examples

input

3
2 2 3
intel
code
ch allenge

output

YES

input

4
1 2 3 1
a
bcdefghi
jklmnopqrstu
vwxyz

output

NO

input

4
13 11 15 15
to be or not to be that is the question
whether tis nobler in the mind to suffer
the slings and arrows of outrageous fortune
or to take arms against a sea of troubles

output

YES

Note

In the first sample, one can split words into syllables in the following way:

in-tel
co-de
ch al-len-ge

Since the word "ch" in the third line doesn't contain vowels, we can ignore it. As the result we get 2 syllabels in first two lines and 3syllables in the third one.

简单的字符计数，一遍过

 #include<iostream>
 #include<string>
 using namespace std;

 int n;
 int P[];
 char S[]={'a','e','i','o','u','y'};
 string s;

 int main()
 {
     cin>>n;
     for(int i=;i<=n;i++)
         cin>>P[i];
     cin.ignore();
     for(int i=;i<=n;i++)
     {
         int num=;
         getline(cin,s);
         for(int j=;j<s.size();j++)
             for(int k=;k<;k++)
                 if(s[j]==S[k])
                 {
                     num++;break;
                 }
         if(num!=P[i])
         {
             cout<<"NO";
             return ;
         }
     }
     cout<<"YES";
     return ;
 }

C. Destroying Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array consisting of n non-negative integers a1, a2, ..., an.

You are going to destroy integers in the array one by one. Thus, you are given the permutation of integers from 1 to n defining the order elements of the array are destroyed.

After each element is destroyed you have to find out the segment of the array, such that it contains no destroyed elements and the sum of its elements is maximum possible. The sum of elements in the empty segment is considered to be 0.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the length of the array.

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).

The third line contains a permutation of integers from 1 to n — the order used to destroy elements.

Output

Print n lines. The i-th line should contain a single integer — the maximum possible sum of elements on the segment containing no destroyed elements, after first i operations are performed.

Examples

input

4
1 3 2 5
3 4 1 2

output

5
4
3
0

input

5
1 2 3 4 5
4 2 3 5 1

output

6
5
5
1
0

input

8
5 5 4 4 6 6 5 5
5 2 8 7 1 3 4 6

output

18
16
11
8
8
6
6
0

Note

Consider the first sample:

1. Third element is destroyed. Array is now 1 3  *  5. Segment with maximum sum 5 consists of one integer 5.
2. Fourth element is destroyed. Array is now 1 3  *   * . Segment with maximum sum 4 consists of two integers 1 3.
3. First element is destroyed. Array is now  *  3  *   * . Segment with maximum sum 3 consists of one integer 3.
4. Last element is destroyed. At this moment there are no valid nonempty segments left in this array, so the answer is equal to 0.

这题仍需优化，超时了。。。

 #include<iostream>
 #include<algorithm>
 using namespace std;

 int n;
 long long T[]={},F[]={},DP[]={};
 long long ans=;

 int main()
 {
     cin>>n;
     for(int i=;i<=n;i++)
     {
         cin>>T[i];
         DP[i]=DP[i-];
         DP[i]+=T[i];
     }
     for(int i=;i<=n;i++)
     {
         cin>>F[i-];
         T[F[i-]]=-;
         sort(F,F+i);
         ans=;
         for(int j=;j<i;j++)
             ans=max(ans,DP[F[j]-]-DP[F[j-]]);
         cout<<max(ans,DP[n]-DP[F[i-]])<<endl;
     }
     return ;
 }