codeforces525B

Pasha and String

 CodeForces - 525B 

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 mdays.

Input

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

Examples

Input

abcdef
1
2

Output

aedcbf

Input

vwxyz
2
2 2

Output

vwxyz

Input

abcdef
3
1 2 3

Output

fbdcea

sol：题意：给出一个字符串，读入ai，表示翻转区间[ai,n-ai+1]
Ps:写splay的大佬可以直接去码splay，不用看下面的XJB做法
易知：每个位置翻转两次就相当于没变，于是处理出每个位置翻转的次数，差分就可以了，然后从1~n/2扫一遍，对于操作次数是奇数的位置与对应的位置交换

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=200005;
int n,Q;
char S[N];
int Sum[N];
int main()
{
    int i;
    scanf("%s",S+1); n=strlen(S+1);
    R(Q);
    for(i=1;i<=Q;i++)
    {
        int x=read();
        Sum[x]++;
        Sum[n-x+2]--;
    }
    for(i=1;i<=n;i++)
    {
        Sum[i]+=Sum[i-1];
    }
    for(i=1;i<=(n/2);i++) if(Sum[i]&1)
    {
        swap(S[i],S[n-i+1]);
    }
    for(i=1;i<=n;i++) putchar(S[i]);
    puts("");
    return 0;
}
/*
input
abcdef
1
2
output
aedcbf

input
vwxyz
2
2 2
output
vwxyz

input
abcdef
3
1 2 3
output
fbdcea

input
keicnqmuqinhsmtudqcilocxkbqgzhbkitmqwttdyoyvcbxincwjryzknubpacsngorexaldfurondbednowemnnlphhboycfavs
2
5 12
output
keiccyobhhphsmtudqcilocxkbqgzhbkitmqwttdyoyvcbxincwjryzknubpacsngorexaldfurondbednowemnnlniqumqnfavs
*/