B. Pasha and String

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Pasha got a very beautiful string s for his birthday, the string consists of lowercase Latin letters. The letters in the string are numbered from 1 to |s| from left to right, where |s| is the length of the given string.

Pasha didn't like his present very much so he decided to change it. After his birthday Pasha spent m days performing the following transformations on his string — each day he chose integer a_i and reversed a piece of string (a segment) from position a_i to position |s| - a_i + 1. It is guaranteed that 2·a_i ≤ |s|.

You face the following task: determine what Pasha's string will look like after m days.

Input

The first line of the input contains Pasha's string s of length from 2 to 2·10⁵ characters, consisting of lowercase Latin letters.

The second line contains a single integer m (1 ≤ m ≤ 10⁵) —  the number of days when Pasha changed his string.

The third line contains m space-separated elements a_i (1 ≤ a_i; 2·a_i ≤ |s|) — the position from which Pasha started transforming the string on the i-th day.

Output

In the first line of the output print what Pasha's string s will look like after m days.

Sample test(s)

Input

abcdef
1
2

Output

aedcbf

Input

vwxyz
2
2 2

Output

vwxyz

Input

abcdef
3
1 2 3

Output

fbdcea

翻转的子串是中心对称的。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <sstream>
 #include <iomanip>
 using namespace std;
 const int INF=0x4fffffff;
 const int EXP=1e-;
 const int MS=;

 char str[*MS];
 int n;
 int flag[MS];

 int main()
 {
       scanf("%s",str);
       scanf("%d",&n);
       memset(flag,,sizeof(flag));
       int len=strlen(str);
       int x;
       for(int i=;i<n;i++)
       {
             scanf("%d",&x);
             flag[x-]++;
       }
       for(int i=;i<len/;i++)
             flag[i]+=flag[i-];
       for(int i=;i<len/;i++)
       {
             if(flag[i]%==)
                   continue;
             else
             {
                   char c=str[i];
                   str[i]=str[len--i];
                   str[len--i]=c;
             }
       }
       printf("%s\n",str);
       return ;
 }