Codeforces 766 C Mahmoud and a Message (基础dp)

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 a_i. 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 = 2and he wants to send string "aaa", he can split it into "a" and "aa" and use 2magical 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 ≤ a_x ≤ 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.

Example

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

Note

In the first example the three ways to split the message are:

• a|a|b
• aa|b
• a|ab

The longest substrings are "aa" and "ab" of length 2.

The minimum number of substrings is 2 in "a|ab" or "aa|b".

Notice that "aab" is not a possible splitting because the letter 'a' appears in a substring of length 3, while a₁ = 2.

题目大意：字符串全部都是小写字母，每一个小写字母都会被赋值，这个值就是这个字母在一个字符串里最多出现的次数。对这个字符串进行拆分成一个个子串，问你有几种情况拆分，然后这些情况中，一个子串里最多有多少个字符，这个字符串最少被拆成几个字符子串。

题目分析：虽说是基础dp，我还是鼓捣了老半天，最后还因为那个数组开小了检查了好久才检查出来，坑爹cf，居然不报内存的错，只说是wrong answer。

   因为每个字母都有只能出现几次的要求，所以要开数组存储下来。

   然后对于这个字符数组遍历，定义len为当前这个字母所能出现的最大次数。然后再循环，用j往后遍历，同时更新len，只要满足i-j+1<=len,那么从i-j+1到i就能划分成一组，所以此时

    dp1[i]=(dp1[i]+dp1[j-1])%mod;dp1数组表示从0到i所能划分的最大组数。

  然后那个一个子串里有多少个字符这个比较简单，直接比较即可。

   最后一个问题则用dp2表示

   dp2[i]=min(dp2[i],dp2[j-1]+1);

   dp1和dp2都需要一定的初始化。

   做完之后感觉这个dp的确比较简单，可惜就是没做出来。。。

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define mod 1000000007using namespace std;const int maxn = 1005;int a[maxn];int dp1[maxn];int dp2[maxn];char s[maxn];int main(){int n;while((scanf("%d",&n))!=EOF){scanf("%s",s+1);for(int i=0;i<26;i++)   scanf("%d",&a[i]);memset(dp1,0,sizeof(dp1));memset(dp2,1,sizeof(dp2));dp1[0]=1;dp2[0]=0;int lmax=0;for(int i=1;i<=n;i++){int len=a[s[i]-'a'];for(int j=i;j>0;j--){len=min(len,a[s[j]-'a']);if(i-j+1>len)  break;lmax=max(i-j+1,lmax);dp1[i]=(dp1[i]+dp1[j-1])%mod;dp2[i]=min(dp2[i],dp2[j-1]+1);}}printf("%d\n%d\n%d\n",dp1[n],lmax,dp2[n]);}return 0;}