C. Mahmoud and a Message

C. Mahmoud and a Message

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Mahmoud wrote a message s of length n. He wants to send it as a birthday present to his friend Moaz who likes strings. He wrote it on a magical paper but he was surprised because some characters disappeared while writing the string. That's because this magical paper doesn't allow character number i in the English alphabet to be written on it in a string of length more than ai. For example, if a1 = 2 he can't write character 'a' on this paper in a string of length 3 or more. String "aa" is allowed while string "aaa" is not.

Mahmoud decided to split the message into some non-empty substrings so that he can write every substring on an independent magical paper and fulfill the condition. The sum of their lengths should be n and they shouldn't overlap. For example, if a1 = 2 and he wants to send string "aaa", he can split it into "a" and "aa" and use 2 magical papers, or into "a", "a" and "a" and use 3 magical papers. He can't split it into "aa" and "aa" because the sum of their lengths is greater than n. He can split the message into single string if it fulfills the conditions.

A substring of string s is a string that consists of some consecutive characters from string s, strings "ab", "abc" and "b" are substrings of string "abc", while strings "acb" and "ac" are not. Any string is a substring of itself.

While Mahmoud was thinking of how to split the message, Ehab told him that there are many ways to split it. After that Mahmoud asked you three questions:

• How many ways are there to split the string into substrings such that every substring fulfills the condition of the magical paper, the sum of their lengths is n and they don't overlap? Compute the answer modulo 109 + 7.
• What is the maximum length of a substring that can appear in some valid splitting?
• What is the minimum number of substrings the message can be spit in?

Two ways are considered different, if the sets of split positions differ. For example, splitting "aa|a" and "a|aa" are considered different splittings of message "aaa".

Input

The first line contains an integer n (1 ≤ n ≤ 103) denoting the length of the message.

The second line contains the message s of length n that consists of lowercase English letters.

The third line contains 26 integers a1, a2, ..., a26 (1 ≤ ax ≤ 103) — the maximum lengths of substring each letter can appear in.

Output

Print three lines.

In the first line print the number of ways to split the message into substrings and fulfill the conditions mentioned in the problem modulo 109  +  7.

In the second line print the length of the longest substring over all the ways.

In the third line print the minimum number of substrings over all the ways.

Examples

input

3aab2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

output

322

input

10abcdeabcde5 5 5 5 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

output

40143

竟然想成了一个区间dp，直接混了，从两边dp去了。其实就是一个很基础的dp。

对于新来的一个字符，新生成的子串自然就是从后面开始长度分别为1.2.3...的一些子串。

所以dp[i]表示长度为i的字符串的总的分隔方法，然后用f[i]表示分隔出的最小的子串数。

j为带有后面新加字符的子串

从i-j+1到i的字符串合法，那么有

1.dp[i]+=dp[i-j];

2.f[i]=min(f[i],f[i-j]+1);

3.如果1~i-j的字符串也合法，那么字符串长度的最大值为max(ans,j);其实前面的必定合法。

因为长度为1的必定合法。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>#include <queue>using namespace std;int n,m;const int MAXN=1e3+7;const int mod=1e9+7;int limit[30];long long dp[MAXN];int f[MAXN];char s[MAXN];int check(int i,int j){    int l=j-i+1;    for(int k=i;k<=j;++k)    {        if(limit[s[k]-'a']<l)return 0;    }    return 1;}int main(){    int i,j;    scanf("%d",&n);    scanf("%s",s+1);    for(i=0;i<26;++i)scanf("%d",&limit[i]);    int MAX=0;    dp[0]=1;    for(i=1;i<=n;++i)//长度为i    {        f[i]=1e9;        for(j=1;j<=i;++j)//长度        {            if(check(i-j+1,i))//后面的当前序列合法            {                dp[i]=(dp[i]+dp[i-j])%mod;                f[i]=min(f[i],f[i-j]+1);                MAX=max(MAX,j);            }        }    }    printf("%I64d\n%d\n%d\n",dp[n],MAX,f[n]);}