There are N squares numbered 1 to N from left to right. Each square has a character written on it, and Square i has a letter s_i. Besides, there is initially one golem on each square.

Snuke cast Q spells to move the golems.

The i-th spell consisted of two characters t_i and d_i, where d_i is `L` or `R`. When Snuke cast this spell, for each square with the character t_i, all golems on that square moved to the square adjacent to the left if d_i is `L`, and moved to the square adjacent to the right if d_i is `R`.

However, when a golem tried to move left from Square 1 or move right from Square N, it disappeared.

Find the number of golems remaining after Snuke cast the Q spells.

Constraints

* 1 \leq N,Q \leq 2 \times 10^{5}
* |s| = N
* s_i and t_i are uppercase English letters.
* d_i is `L` or `R`.

Input

Input is given from Standard Input in the following format:


N Q
s
t_1 d_1
\vdots
t_{Q} d_Q


Output

Print the answer.

Examples

Input

3 4
ABC
A L
B L
B R
A R


Output

2


Input

8 3
AABCBDBA
A L
B R
A R


Output

5


Input

10 15
SNCZWRCEWB
B R
R R
E R
W R
Z L
S R
Q L
W L
B R
C L
A L
N L
E R
Z L
S L


Output

3
#include <bits/stdc++.h>
using namespace std;

int N,Q;
string s;
vector<char> td;

bool vanish_l(int x){
	int pos=x;
	for(int i=0;i<Q;i++){
		char t=td[2*i],d=td[2*i+1];
		if(s[pos]==t){
			if(d=='L')pos--;
			else pos++;
		}
		if(pos==-1)return true;
		if(pos==N)return false;
	}
	return false;
}

bool vanish_r(int x){
	int pos=x;
	for(int i=0;i<Q;i++){
		char t=td[2*i],d=td[2*i+1];
		if(s[pos]==t){
			if(d=='L')pos--;
			else pos++;
		}
		if(pos==-1)return false;
		if(pos==N)return true;
	}
	return false;
}

int main(){
	cin >> N >> Q >> s;
	for(int i=0;i<2*Q;i++){
		char tmp;
		cin >> tmp;
		td.push_back(tmp);
	}
	
	int left,right;
	
	int ok,ng;
	
	ok=-1;
	ng=N;
	while(abs(ng-ok)>1){
		int mid=(ok+ng)/2;
		if(vanish_l(mid))ok=mid;
		else ng=mid;
	}
	left=ok;
	
	ok=N;
	ng=-1;
	while(abs(ng-ok)>1){
		int mid=(ok+ng)/2;
		if(vanish_r(mid))ok=mid;
		else ng=mid;
	}
	right=ok;
	
	cout << right-left-1 << endl;
	
	return 0;
}