CodeForces

A string a of length m is called antipalindromic iff m is even, and for each i (1 ≤ i ≤ m) a_i ≠ a_{m - i + 1}.

Ivan has a string s consisting of n lowercase Latin letters; n is even. He wants to form some string t that will be an antipalindromic permutation of s. Also Ivan has denoted the beauty of index i as b_i, and the beauty of t as the sum of b_i among all indices i such that s_i = t_i.

Help Ivan to determine maximum possible beauty of t he can get.

Input

The first line contains one integer n (2 ≤ n ≤ 100, n is even) — the number of characters in s.

The second line contains the string s itself. It consists of only lowercase Latin letters, and it is guaranteed that its letters can be reordered to form an antipalindromic string.

The third line contains n integer numbers b₁, b₂, ..., b_n (1 ≤ b_i ≤ 100), where b_i is the beauty of index i.

Output

Print one number — the maximum possible beauty of t.

Examples

Input

8
abacabac
1 1 1 1 1 1 1 1

Output

Input

8
abaccaba
1 2 3 4 5 6 7 8

Output

Input

8
abacabca
1 2 3 4 4 3 2 1

Output

题意：给定长长度为N的字符串，现在求一个重排列，使得对称位置不相同。如果重排后i位置的字母和原来相同，就得到对应位置的得分，求最大得分。

保证N是偶数，保证有满足条件的重排。

思路：最大费用最大流。

建图：

S-->26个字母:（字母个数，0）；

字母-->N/2个位置：（1，val）；val尽可能大就行，三种情况讨论。

N/2个位置-->T:（2，0）

#include<bits/stdc++.h>
using namespace std;
const int maxn=100000;
const int inf=1<<30;int To[maxn],Laxt[maxn],Next[maxn],cap[maxn],cost[maxn];
int S,T,cnt=1,dis[maxn],ans;
bool inq[maxn],vis[maxn];
deque<int>q;
void add(int u,int v,int c,int cc)
{ 
    Next[++cnt]=Laxt[u];Laxt[u]=cnt; To[cnt]=v;cap[cnt]=c;cost[cnt]=-cc; 
    Next[++cnt]=Laxt[v];Laxt[v]=cnt; To[cnt]=u;cap[cnt]=0;cost[cnt]=cc; 
}
bool spfa()
{
    for(int i=0;i<=T;i++) inq[i]=0;
    for(int i=0;i<=T;i++) dis[i]=inf;
    inq[T]=1; dis[T]=0; q.push_back(T);
    while(!q.empty())
    {    
        int u=q.front(); q.pop_front();
        inq[u]=0;
        for(int i=Laxt[u];i;i=Next[i])
        {
            int v=To[i];
            if(cap[i^1]&&dis[v]>dis[u]-cost[i])
            {
                dis[v]=dis[u]-cost[i];
                if(!inq[u]){
                    inq[v]=1;
                    if(q.empty()||dis[v]>dis[q.front()]) q.push_back(v);
                    else q.push_front(v);
                }
            }
        }
    }
    return dis[S]<inf;
}
int dfs(int u,int flow)
{
    vis[u]=1;
    if(u==T||flow==0) return flow;
    int tmp,delta=0;
    for(int i=Laxt[u];i;i=Next[i])
    {
        int v=To[i];
        if((!vis[v])&&cap[i]&&dis[v]==dis[u]-cost[i])
        {
            tmp=dfs(v,min(cap[i],flow-delta));
            delta+=tmp; cap[i]-=tmp; cap[i^1]+=tmp;
        }
    }
    return delta;
}
char c[maxn]; int v[maxn],num[maxn];
int main()
{
    int N,i,j;
    scanf("%d",&N); scanf("%s",c+1);
    for(i=1;i<=N;i++) scanf("%d",&v[i]),num[c[i]-'a'+1]++;
    S=0; T=26+N/2+1;
    for(i=1;i<=26;i++) add(S,i,num[i],0);
    for(i=1;i<=26;i++)
     for(j=1;j<=N/2;j++){
         if(c[j]-'a'+1==i&&c[N+1-j]-'a'+1==i) add(i,j+26,1,max(v[j],v[N+1-j]));
         else if(c[j]-'a'+1==i) add(i,j+26,1,v[j]);
         else if(c[N+1-j]-'a'+1==i) add(i,j+26,1,v[N+1-j]);
         else add(i,j+26,1,0);
    }
    for(i=1;i<=N/2;i++) add(i+26,T,2,0);
    while(spfa()){
        vis[T]=1;
        while(vis[T]){
            for(i=0;i<=T;i++) vis[i]=0;
            ans-=dis[S]*dfs(S,N);
        }
    }
    printf("%d
",ans);
    return 0;
}