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>
#include<vector>
#include<algorithm>
#include<string>
#include<iostream>
using namespace std;
typedef long long ll;
ll mod = 1000000007;
ll dp[3002][3002];
int dat[3002];
int rui[3002];
int main()
{
	int num, way;
	scanf("%d%d", &num, &way);
	string s;
	cin >> s;
	reverse(s.begin(), s.end());
	s.push_back('0');
	reverse(s.begin(), s.end());
	num++;
	for (int i = 1; i < num; i++)rui[i] = rui[i - 1] + s[i] - '0';
	for (int i = 0; i < num; i++)dat[i] = i;
	for (int i = 0; i < way; i++)
	{
		int za, zb;
		scanf("%d%d", &za, &zb);
		dat[za] = max(dat[za], zb);
	}
	for (int i = 1; i < num; i++)dat[i] = max(dat[i], dat[i - 1]);
	dp[0][0] = 1;
	for (int i = 1; i < num; i++)
	{
		int a0 = (dat[i] - dat[i - 1]) - (rui[dat[i]] - rui[dat[i - 1]]);
		int a1 = (rui[dat[i]] - rui[dat[i - 1]]);
		int sum = dat[i - 1] - (i - 1);
		//printf("%d %d %d\n", a0, a1, sum);
		for (int j = 0; j <= sum; j++)
		{
			if (sum - j + a0 != 0)dp[i][j + a1] += dp[i - 1][j], dp[i][j + a1] %= mod;
			if (j + a1 != 0)dp[i][j + a1 - 1] += dp[i - 1][j], dp[i][j + a1 - 1] %= mod;
		}
		//for (int j = 0; j < num; j++)printf("%lld ", dp[i][j]); printf("\n");
	}
	printf("%lld\n", dp[num - 1][0]);
}