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;

using ll = long long;

const ll MOD = 1e9 + 7;

ll dp[3005][3005], comb[3005][3005];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	string s;
	cin >> n >> m >> s;

	vector<int> l, r;
	int rmax = 0;
	for (int i = 0; i < m; i++) {
		int tl, tr;
		cin >> tl >> tr;
		tl--;
		rmax = max(rmax, tr);
		l.push_back(tl);
		r.push_back(rmax);
	}

	vector<int> sum(n + 1, 0);
	for (int i = 0; i < n; i++) sum[i + 1] = sum[i] + s[i] - '0';

	for (int i = 0; i <= n; i++) comb[i][0] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= i; j++) {
			comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % MOD;
		}
	}
	
	dp[0][sum[l[0]]] = 1;
	l.push_back(n);
	for (int i = 0; i < m; i++) {
		if (i > 0 && l[i] == l[i - 1] && r[i] < r[i - 1]) {
			for (int j = 0; j <= n; j++) {
				dp[i + 1][j] = dp[i][j];
			}
		}
		for (int j = 0; j <= sum[r[i]]; j++) {
			if (dp[i][j] == 0) continue;
			int rest = sum[r[i]] - j;
			int plus = max(sum[l[i + 1]] - sum[r[i]], 0);
			int L = min(r[i], l[i + 1]) - l[i];
			for (int k = max(L - (r[i] - l[i] - rest), 0); k <= min(L, rest); k++) {
				(dp[i + 1][j + k + plus] += dp[i][j] * comb[L][k] % MOD) %= MOD;
			}
		}
	}

	cout << dp[m][sum[n]] << endl;
	return 0;
}