【Codeforces Round 299 (Div 2)D】【KMP 本质是最前与最后匹配】Tavas and Malekas 长度为n的匹配串被模板串多位点覆盖的匹配串个数

D. Tavas and Malekas

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Tavas is a strange creature. Usually "zzz" comes out of people's mouth while sleeping, but string s of length n comes out from Tavas' mouth instead.

Today Tavas fell asleep in Malekas' place. While he was sleeping, Malekas did a little process on s. Malekas has a favorite string p. He determined all positions x₁ < x₂ < ... < x_k where p matches s. More formally, for each x_i (1 ≤ i ≤ k) he condition s_xis_xi + 1... s_{xi + |p| - 1} = pis fullfilled.

Then Malekas wrote down one of subsequences of x₁, x₂, ... x_k (possibly, he didn't write anything) on a piece of paper. Here a sequence b is a subsequence of sequence a if and only if we can turn a into b by removing some of its elements (maybe no one of them or all).

After Tavas woke up, Malekas told him everything. He couldn't remember string s, but he knew that both p and s only contains lowercase English letters and also he had the subsequence he had written on that piece of paper.

Tavas wonders, what is the number of possible values of s? He asked SaDDas, but he wasn't smart enough to solve this. So, Tavas asked you to calculate this number for him.

Answer can be very large, so Tavas wants you to print the answer modulo 10⁹ + 7.

Input

The first line contains two integers n and m, the length of s and the length of the subsequence Malekas wrote down (1 ≤ n ≤ 10⁶ and0 ≤ m ≤ n - |p| + 1).

The second line contains string p (1 ≤ |p| ≤ n).

The next line contains m space separated integers y₁, y₂, ..., y_m, Malekas' subsequence (1 ≤ y₁ < y₂ < ... < y_m ≤ n - |p| + 1).

Output

In a single line print the answer modulo 1000 000 007.

Sample test(s)

input

6 2ioi1 3

output

26

input

5 2ioi1 2

output

0

Note

In the first sample test all strings of form "ioioi?" where the question mark replaces arbitrary English letter satisfy.

Here |x| denotes the length of string x.

Please note that it's possible that there is no such string (answer is 0).

#include<stdio.h>#include<string.h>#include<ctype.h>#include<math.h>#include<iostream>#include<string>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T> inline void gmax(T &a,T b){if(b>a)a=b;}template <class T> inline void gmin(T &a,T b){if(b<a)a=b;}const int N=1e6+10,M=0,Z=1e9+7,ms63=1061109567;int n,m,g;char s[N];int nxt[N];int p[N];bool e[N];void getnxt(){int j=0;nxt[1]=0;for(int i=2;i<=m;++i){while(j&&s[j+1]!=s[i])j=nxt[j];if(s[j+1]==s[i])++j;nxt[i]=j;}MS(e,0);for(int p=m;p;p=nxt[p])e[p]=1;}int solve(){for(int i=g;i>=1;--i){int dis=p[i+1]-p[i];if(dis>=m)n-=m;else{int match=m-dis;if(!e[match])return 0;else n-=dis;}}LL ans=1;while(n--)ans=ans*26%Z;return ans;}int main(){while(~scanf("%d%d",&n,&g)){scanf("%s",s+1);m=strlen(s+1);getnxt();for(int i=1;i<=g;++i)scanf("%d",&p[i]);p[g+1]=n+1;printf("%d\n",solve());}return 0;}/*【题意】给定一个字符串，作为匹配串但是只长度为n（1<=n<=1e6），（所有字符仅为'a'~'z')。在这个字符串中，已知给定的子串s至少出现了g次（0<=g<=n-|s|+1），并且给你s出现的g次分别是在匹配串的什么位置。问你匹配串有多少种构成方式。【类型】KMP【分析】难度1：如果g=0，显然答案是26^n。难度2：如果任意两个子串的头位置相差都至少是m，显然答案是26^(n-m*g)难度3：现在子串之间产生了位置冲突。也就是"子串a,子串b"，子串a的尾巴与子串b的头部叠在了一起。而事实上，子串a=子串b=串s。假设这两个串的头端点相差了dis，重叠部分的长度显然就是m-dis，设为match。那么我们其实只需要验证，串s的最前match位与最后match位是否相同。天哪，这就是KMP所可以简单实现的功能！对s串做KMP，从尾巴开始沿着fail指针nxt向上跳，设一个能跳到的位置为pos，基于KMP，那么就有，串s的最前pos位与最后pos位相同，我们打上标记即可。我们查看标记，如果发现其不满足题目要求，那么答案就是0，否则我们就有dis长度的串也是固定的。用总的长度n减去。这样最后的答案就是26^n，现在这个n表示没有被s覆盖过、决定过的长度。*/