Educational Codeforces Round 36 (Rated for Div. 2)

A. Garden

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Luba thinks about watering her garden. The garden can be represented as a segment of length k. Luba has got n buckets, the i-th bucket allows her to water some continuous subsegment of garden of length exactly ai each hour. Luba can't water any parts of the garden that were already watered, also she can't water the ground outside the garden.

Luba has to choose one of the buckets in order to water the garden as fast as possible (as mentioned above, each hour she will water some continuous subsegment of length ai if she chooses the i-th bucket). Help her to determine the minimum number of hours she has to spend watering the garden. It is guaranteed that Luba can always choose a bucket so it is possible water the garden.

See the examples for better understanding.

Input

The first line of input contains two integer numbers n and k (1 ≤ n, k ≤ 100) — the number of buckets and the length of the garden, respectively.

The second line of input contains n integer numbers ai (1 ≤ ai ≤ 100) — the length of the segment that can be watered by the i-th bucket in one hour.

It is guaranteed that there is at least one bucket such that it is possible to water the garden in integer number of hours using only this bucket.

Output

Print one integer number — the minimum number of hours required to water the garden.

Examples

input

Copy

3 6
2 3 5

output

2

input

Copy

6 7
1 2 3 4 5 6

output

7

Note

In the first test the best option is to choose the bucket that allows to water the segment of length 3. We can't choose the bucket that allows to water the segment of length 5 because then we can't water the whole garden.

In the second test we can choose only the bucket that allows us to water the segment of length 1.

k能整除当前数的时候，找出那个最大的倍数

#include <bits/stdc++.h>
using namespace std;
int a[];
int main()
{
    int n,k,f=;
    cin>>n>>k;
    for(int i=,x;i<n;i++)
    {
        cin>>x;
        if(k%x==&&x>f)f=x;
    }
    cout<<k/f;
    return ;
}

B. Browser

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Luba is surfing the Internet. She currently has n opened tabs in her browser, indexed from 1 to n from left to right. The mouse cursor is currently located at the pos-th tab. Luba needs to use the tabs with indices from l to r (inclusive) for her studies, and she wants to close all the tabs that don't belong to this segment as fast as possible.

Each second Luba can either try moving the cursor to the left or to the right (if the cursor is currently at the tab i, then she can move it to the tab max(i - 1, a) or to the tab min(i + 1, b)) or try closing all the tabs to the left or to the right of the cursor (if the cursor is currently at the tab i, she can close all the tabs with indices from segment [a, i - 1] or from segment [i + 1, b]). In the aforementioned expressions a and b denote the minimum and maximum index of an unclosed tab, respectively. For example, if there were 7 tabs initially and tabs 1, 2 and 7are closed, then a = 3, b = 6.

What is the minimum number of seconds Luba has to spend in order to leave only the tabs with initial indices from l to r inclusiveopened?

Input

The only line of input contains four integer numbers n, pos, l, r (1 ≤ n ≤ 100, 1 ≤ pos ≤ n, 1 ≤ l ≤ r ≤ n) — the number of the tabs, the cursor position and the segment which Luba needs to leave opened.

Output

Print one integer equal to the minimum number of seconds required to close all the tabs outside the segment [l, r].

Examples

input

Copy

6 3 2 4

output

5

input

Copy

6 3 1 3

output

1

input

Copy

5 2 1 5

output

0

Note

In the first test Luba can do the following operations: shift the mouse cursor to the tab 2, close all the tabs to the left of it, shift the mouse cursor to the tab 3, then to the tab 4, and then close all the tabs to the right of it.

In the second test she only needs to close all the tabs to the right of the current position of the cursor.

In the third test Luba doesn't need to do anything.

给你n pos l r，分别代表几个网页，你现在在第几个网页，你需要关闭l左边的网页，和r右边的网页，你可以移到一个网页把之前或之后的关闭，问你最少走几步

那么我枚举到l r的距离即可分别选择先向左还是先向右

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,pos,l,r,f=;
    cin>>n>>pos>>l>>r;
    if(l==&&r==n)
        f=;
    else if(l==)
        f=abs(pos-r)+;
    else if(r==n)
        f=abs(pos-l)+;
    else
    {
        if(pos<l)
            f=r-pos+;
        else if(pos>r)
            f=pos-l+;
        else
        f=min(pos-l,r-pos)+r-l+;
    }
    cout<<f;
    return ;
}

C. Permute Digits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two positive integer numbers a and b. Permute (change order) of the digits of a to construct maximal number not exceeding b. No number in input and/or output can start with the digit 0.

It is allowed to leave a as it is.

Input

The first line contains integer a (1 ≤ a ≤ 1018). The second line contains integer b (1 ≤ b ≤ 1018). Numbers don't have leading zeroes. It is guaranteed that answer exists.

Output

Print the maximum possible number that is a permutation of digits of a and is not greater than b. The answer can't have any leading zeroes. It is guaranteed that the answer exists.

The number in the output should have exactly the same length as number a. It should be a permutation of digits of a.

Examples

input

Copy

123
222

output

213

input

Copy

3921
10000

output

9321

input

Copy

4940
5000

output

4940

给你一个数x和n，你可以把m随意更改顺序构造不大于n的最大的数

当时我错的条件就是等于

#include <bits/stdc++.h>
using namespace std;
string s,c;
int cmp(char x,char y)
{
    return x>y;
}
long long ans=,b;
void dfs(int f,long long x,string a)
{
    if(c[f])
    {
        long long x1=x;
        string bb=a;
        int mi=,mif=-,maf=-;
        for(int i=; a[i]; i++)
            if(a[i]==c[f])maf=i;
            else if(a[i]<c[f]&&a[i]>=mi)mif=i,mi=a[i];
        if(mif!=-)
        {
            if(f==&&a[mif]=='')
            {

            }
            else
            {
                x=x*+a[mif]-'';
                a[mif]=;
                sort(a.begin(),a.end(),cmp);
                for(int i=;i<(int)a.size()-f-;i++)
                x=x*+a[i]-'';
                if(x<=b)ans=max(ans,x);
            }
        }
        if(maf!=-)
        {
            x1=x1*+bb[maf]-'';
            bb[maf]=;
            dfs(f+,x1,bb);
        }
    }
    else
    {
        if(x<=b)ans=max(ans,x);
        return;
    }
}
int main()
{
    long long a;
    cin>>a>>b;
    stringstream ss;
    ss<<a;
    s=ss.str();
    stringstream sss;
    sss<<b;
    c=sss.str();
    if(s.size()<c.size())
    {
        sort(s.begin(),s.end(),cmp);
        cout<<s;
    }
    else
    {
        dfs(,,s);
        cout<<ans;
    }
    return ;
}