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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

const ll M = 1000000007;
char str[3010];
int n, m, c = 1, sum[3010], s[3010], e[3010];
ll dp[3010][3010], cm[3010][3010], chk[3010][3010];

int get(int s, int e){ return sum[e] - sum[s - 1]; }

ll f(int x, int y){
	if(x > m) return 1;
	if(y < 0 || e[x] - s[x] + 1 < y) return 0;
	if(chk[x][y]) return dp[x][y];
	if(e[x] < s[x + 1]){
		dp[x][y] = cm[e[x] - s[x] + 1][y] * f(x + 1, get(s[x + 1], e[x + 1])) % M;
	}
	else{
		for(int i = max(0, y - (e[x] - s[x + 1] + 1)); i <= min(s[x + 1] - s[x], y); i++){
			dp[x][y] = (dp[x][y] + cm[s[x + 1] - s[x]][i] *
						f(x + 1, y - i + get(e[x] + 1, e[x + 1]))) % M;
		}
	}
	chk[x][y] = 1;
	return dp[x][y];
}

int main(){
	scanf("%d%d%s", &n, &m, str + 1);
	for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (str[i] == '1');
	for(int x, y, l = 0; m--; ){
		scanf("%d%d", &x, &y);
		if(y <= l) continue;
		l = y;
		if(s[c - 1] == x) e[c - 1] = y;
		else{
			s[c] = x; e[c] = y;
			c++;
		}
	}
	m = c - 1;
	s[m + 1] = n + 1;
	for(int i = 0; i <= n; i++){
		cm[i][0] = 1;
		for(int j = 1; j <= i; j++) cm[i][j] = (cm[i - 1][j - 1] + cm[i - 1][j]) % M;
	}
	printf("%lld\n", f(1, get(s[1], e[1])));
}