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>
#define int long long 
using namespace std;

const int N = 3000 + 7, mod = 1e9 + 7;

int c[N], l[N], r[N], n, m, dp[N][N];

string str;


int f(int idx, int one) {
	if(idx > n) {
		return 1;
	}
	if(~dp[idx][one]) {
		return dp[idx][one];
	}
	int res = 0;
	if(r[idx] - idx >= one) {
		if(l[idx + 1] != l[idx]) { 
			res += f(idx + 1, one + c[r[idx + 1]] - c[r[idx]]); 
			res %= mod;
		} else {
			res += f(idx + 1, one);
			res %= mod;
		}
	}

	if(one) {
		if(l[idx + 1] != l[idx]) {
			res += f(idx + 1, one - 1 + c[r[idx + 1]] - c[r[idx]]);
			res %= mod;
		} else {
			res += f(idx + 1, one - 1);
			res %= mod;
		}
	}
	
	return dp[idx][one] = res;
}

int32_t main() {
	ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);

	cin >> n >> m >> str;
	for(int i = 1; i <= m; i++) {
		int lx, rx; cin >> lx >> rx;
		r[lx] = max(rx, r[lx]);
	}

	for(int i = 1; i <= n; i++) {
		r[i] = max(i, r[i]);
		l[i] = r[l[i - 1]] >= r[i] ? l[i - 1] : i;
		r[i] = r[l[i]];
		c[i] = c[i - 1] + (str[i - 1] == '1');
	}

	c[n + 1] = c[n];

	l[n + 1] = r[n + 1] = n + 1;

	memset(dp, -1, sizeof dp);
	cout << f(1, c[r[1]]) << endl;
}