CF1096:D. Easy Problem(DP)

Vasya is preparing a contest, and now he has written a statement for an easy problem. The statement is a string of length n

consisting of lowercase Latin latters. Vasya thinks that the statement can be considered hard if it contains a subsequence hard; otherwise the statement is easy. For example, hard, hzazrzd, haaaaard can be considered hard statements, while har, hart and drah are easy statements.

Vasya doesn't want the statement to be hard. He may remove some characters from the statement in order to make it easy. But, of course, some parts of the statement can be crucial to understanding. Initially the ambiguity of the statement is 0

, and removing i-th character increases the ambiguity by ai (the index of each character is considered as it was in the original statement, so, for example, if you delete character r from hard, and then character d, the index of d is still 4

even though you delete it from the string had).

Vasya wants to calculate the minimum ambiguity of the statement, if he removes some characters (possibly zero) so that the statement is easy. Help him to do it!

Recall that subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

Input

The first line contains one integer n

(1≤n≤105

) — the length of the statement.

The second line contains one string s

of length n

, consisting of lowercase Latin letters — the statement written by Vasya.

The third line contains n

integers a1,a2,…,an (1≤ai≤998244353

).

Output

Print minimum possible ambiguity of the statement after Vasya deletes some (possibly zero) characters so the resulting statement is easy.

Examples

Input

Copy

6
hhardh
3 2 9 11 7 1

Output

Copy

5

Input

Copy

8
hhzarwde
3 2 6 9 4 8 7 1

Output

Copy

4

Input

Copy

6
hhaarr
1 2 3 4 5 6

Output

Copy

0

Note

In the first example, first two characters are removed so the result is ardh.

In the second example, 5

-th character is removed so the result is hhzawde.

In the third example there's no need to remove anything.

題意：給定一個字符串，每個字符自帶權值，讓你刪去一些，使得不存在子序列“hard”，問最下的權值是多少。

思路：因為有順序問題，所以我們記錄維護到當前最長的前綴的代價。1對應h，2對應ha，3對應har，4對應hard，然后就不難寫出方程了。

（復雜度O(5N)，比賽時寫了個2進制，復雜度O(16N)；傻了

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
#define ll long long
const int maxn=200010;
int a[maxn];ll dp[maxn][5],ans; char c[maxn]; int Laxt[maxn];
int id(char s){
    if(s=='h') return 1;
    if(s=='a') return 2;
    if(s=='r') return 3;
    if(s=='d') return 4;
    return -1;
}
void ADD(ll &x,ll y){
    if(y==-1) return ;
    if(x==-1) x=y;
    else x=min(x,y);
}
int main()
{
    int N;
    scanf("%d%s",&N,c+1); memset(dp,-1,sizeof(dp));
    rep(i,1,N) scanf("%d",&a[i]);
    dp[0][0]=0;
    rep(i,1,N){
        int p=id(c[i]);
        if(p==-1) {
            rep(j,0,4) dp[i][j]=dp[i-1][j];
            continue;
        }
        if(dp[i-1][p-1]!=-1) ADD(dp[i][p-1],dp[i-1][p-1]+a[i]);
        ADD(dp[i][p],dp[i-1][p-1]);
        rep(j,0,4){
            if(j==p-1) continue;
            ADD(dp[i][j],dp[i-1][j]);
        }
    }
    ans=1LL<<60; rep(i,0,3) if(dp[N][i]!=-1) ans=min(ans,dp[N][i]);
    printf("%lld\n",ans);
    return 0;
}