There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143
#include <stdio.h>

const long long mod = 1000000007;

int N, M;
char S[3030];
int l[3030], r[3030];
long long comb[3030][3030];
long long p[3030],n[3030];

int main()
{
	scanf("%d %d %s", &N, &M, S);
	for (int i = 0; i < M; i++){
		scanf("%d %d", &l[i], &r[i]);
		l[i]--; r[i]--;
		if (i > 0 && l[i - 1] == l[i])
		{
			if (r[i - 1] < r[i])
				r[i - 1] = r[i];
			i--;
			M--;
		}
		else if (i > 0 && r[i] <= r[i-1])
		{
			i--;
			M--;
		}
	}

	for (int i = 0; i <= N; i++){
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; j++) comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % mod;
	}

	int have = 0;
	p[0] = 1;
	for (int i = 0; i < M; i++){
		int nxt = ((i == M - 1) ? 0 : (r[i] < l[i + 1] ? 0 : r[i] - l[i + 1] + 1));

		int L = (i == 0 || r[i - 1] < l[i]) ? l[i] : r[i - 1] + 1;
		int R = r[i];

		int cnt = 0;
		for (int j = L; j <= R; j++) if (S[j] == '1') cnt++;
		int all = r[i] - l[i] + 1;

		for (int k = 0; k <= nxt; k++) n[k] = 0;

		for (int j = 0; j <= have; j++){
			int v = cnt + j;
			for (int k = 0; k <= nxt; k++){
				int p1 = v - k;
				int p0 = (all - nxt) - p1;
				if (p1 < 0 || p0 < 0) continue;

				n[k] = (n[k] + p[j] * comb[all - nxt][p1]) % mod;
			}
		}

		for (int k = 0; k <= nxt; k++) p[k] = n[k];
		have = nxt;
	}

	printf("%lld\n", p[0]);

	return 0;
}