Recently, Norge found a string s=s1s2…sns=s1s2…sn consisting of nn lowercase Latin letters. As an exercise to improve his typing speed, he decided to type all substrings of the string ss. Yes, all n(n+1)2n(n+1)2 of them!

A substring of ss is a non-empty string x=s[a…b]=sasa+1…sbx=s[a…b]=sasa+1…sb (1≤a≤b≤n1≤a≤b≤n). For example, “auto” and “ton” are substrings of “automaton”.

Shortly after the start of the exercise, Norge realized that his keyboard was broken, namely, he could use only kk Latin letters c1,c2,…,ckc1,c2,…,ck out of 2626.

After that, Norge became interested in how many substrings of the string ss he could still type using his broken keyboard. Help him to find this number.

Input
The first line contains two space-separated integers nn and kk (1≤n≤2?1051≤n≤2?105, 1≤k≤261≤k≤26) — the length of the string ss and the number of Latin letters still available on the keyboard.

The second line contains the string ss consisting of exactly nn lowercase Latin letters.

The third line contains kk space-separated distinct lowercase Latin letters c1,c2,…,ckc1,c2,…,ck — the letters still available on the keyboard.

Output
Print a single number — the number of substrings of ss that can be typed using only available letters c1,c2,…,ckc1,c2,…,ck.

Examples
Input
7 2
abacaba
a b
Output
12
Input
10 3
sadfaasdda
f a d
Output
21
Input
7 1
aaaaaaa
b
Output
0
Note
In the first example Norge can print substrings s[1…2]s[1…2], s[2…3]s[2…3], s[1…3]s[1…3], s[1…1]s[1…1], s[2…2]s[2…2], s[3…3]s[3…3], s[5…6]s[5…6], s[6…7]s[6…7], s[5…7]s[5…7], s[5…5]s[5…5], s[6…6]s[6…6], s[7…7]s[7…7].
比較水的一道題目，主要就是考察一個(gè)字符串的子串有多少個(gè)。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;vector<int> p; string s; int n,m;inline void dfs(int &i,ll &cnt) {if(i==n) return ;int pos=lower_bound(p.begin(),p.end(),s[i]-'a')-p.begin();if(p[pos]!=s[i]-'a') return ;dfs(i+=1,cnt+=1); } int main() {scanf("%d%d",&n,&m);cin>>s;char c;for(int i=0;i<m;i++) cin>>c,p.push_back(c-'a');//for(int i=0;i<p.size();i++) cout<<p[i]<<" ";cout<<endl;sort(p.begin(),p.end());ll ans=0;ll cnt=0;for(int i=0;i<n;){int pos=lower_bound(p.begin(),p.end(),s[i]-'a')-p.begin();if(p[pos]==(s[i]-'a')){cnt++;i++;dfs(i,cnt);ans+=(cnt*(cnt+1))/2;}else cnt=0,i++;}cout<<ans<<endl;return 0; }

努力加油a啊，(^o)/~

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