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> t(q), d(q);

int solve(int pos){
	for(int i = 0; i < q; ++i){
		if(s[pos] == t[i]){
			if(d[i]=='R'){
				pos++;
			}else{
				pos--;
			}
		}
		if(pos == -1) return -1;
		if(pos == n) return 1;
	}
	return 0;
}

int main(void){
	cin >> n >> q >> s;
	t.resize(q); d.resize(q);
	for(int i = 0; i < q; ++i){
		cin >> t[i] >> d[i];
	}
	int l, r;

	/* 
	 * Meguru's binary search
	 * [ok, ng)
	 */
	int ok = 0;
	int ng = n-1;
	while(abs(ok-ng)>1){
		int mid = (ok + ng) / 2;
		if(solve(mid) == -1){
			ok = mid;
		}else{
			ng = mid;
		}
	}
	l = ok; if(!solve(l)) l++;

	/* 
	 * Meguru's binary search 
	 * (ng,ok]
	 */
	ok = n-1;
	ng = 0;
	while(abs(ok-ng)>1){
		int mid = (ok + ng) / 2;
		if(solve(mid) == 1){
			ok = mid;
		}else{
			ng = mid;
		}
	}
	r = ok; if(!solve(r)) r++;

	//cout << l << " " << r << endl;
	cout << r-l-1 << endl;
	return 0;
}
