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 <iostream>
#include <vector>

using namespace std;

int N, Q;
string str;
vector<char> t, d;

int is_down(int k) {
	int pos = k;
	for (int i = 0; i < Q; i++) {
		if (pos == -1 || pos == N) return pos;
		if (str[pos] == t[i]) {
			if (d[i] == 'L') pos--;
			else pos++;
		}
	}
	return pos;
}


int main() {

	cin >> N >> Q;
	cin >> str;
	
	for (int i = 0; i < Q; i++)
	{
		char tmp1, tmp2;
		cin >> tmp1 >> tmp2;
		t.push_back(tmp1);
		d.push_back(tmp2);
	}

	//左に落ちないゴーレムのうち、最も左端のものを二分探索
	int left= -1, right = N;
	while (left +1 < right) {
		int mid = (left + right) / 2;
		if (is_down(mid) == -1) left = mid;
		else right = mid;
	}
	int remainL = right;

	//右に落ちないゴーレムのうち、最も右端のものを二分探索
	left = -1, right = N;
	while (left + 1 < right) {
		int mid = (left + right) / 2;
		if (is_down(mid) == N) right = mid;
		else left = mid;
	}
	int remainR = left;
	int ans = remainR - remainL + 1;

	cout << ans << endl;

	return 0;
}