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;

const int N = 200000 + 5;

char str[N];
struct Node{
    char c[2], d[2];
}node[N];

int main ()
{
    int n, q;
    cin >> n >> q >> str + 1;
    for (int i = 1; i <= q; ++i) cin >> node[i].c >> node[i].d;
    int ans1 = 0, ans2 = 0;
    for (int i = q; i >= 1; --i) {
        if (ans1 != n) {
            if (node[i].c[0] == str[ans1 + 1] && node[i].d[0] == 'L') {
                ++ans1;
            }
            else if (ans1 && node[i].c[0] == str[ans1] && node[i].d[0] == 'R'){
                --ans1;
            }
        }
        if (ans2 != n) {
            if (node[i].c[0] == str[n - ans2] && node[i].d[0] == 'R') {
                ++ans2;
            }
            else if (ans2 && node[i].c[0] == str[n - ans2 + 1] && node[i].d[0] == 'L') {
                --ans2;
            }
        }
    }
    cout << n - min(n, ans1 + ans2) << endl;
    return 0;
}
