/*

     题意：给出m个位置，每次把[p,len-p+1]内的字符子串反转，输出最后的结果

     字符串处理：朴素的方法超时，想到结果要么是反转要么没有反转，所以记录

                 每个转换的次数，把每次要反转的反转就不超时了:)

 */

 #include <cstdio>

 #include <algorithm>

 #include <cstring>

 using namespace std;

 const int MAXN = 2e5 + ;

 const int INF = 0x3f3f3f3f;

 char s[MAXN];

 int to[MAXN];

 int main(void)        //Codeforces Round #297 (Div. 2) B. Pasha and String

 {

     int m;

     scanf ("%s", s + );

     int len = strlen (s + );

     scanf ("%d", &m);

     while (m--)

     {

         int p;    scanf ("%d", &p);

         int q = len - p + ;

         to[p]++;

     }

     int sum = ;

     for (int i=; i<=len/; ++i)

     {

         sum += to[i];

         if (sum & )    swap (s[i], s[len-i+]);

     }

     printf ("%s", s + );

     return ;

 }

 /*

 abcdef

 1

 2

 vwxyz

 2

 2 2

 abcdef

 3

 1 2 3

 */