Problem: 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

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>

using namespace std;

const int MAXN = 3000 + 100;
const int MOD = 1000000000 + 7;

int n, m;
char s[MAXN];
int sum[MAXN];
int f[MAXN][MAXN], rig[MAXN];

void add(int &x, int y)
{
	x = (x + y) % MOD;
}

int main()
{
	ios::sync_with_stdio(false);
	// freopen("1.in", "r", stdin);
	// freopen("1.out", "w", stdout);

	cin >> n >> m;
	cin >> (s + 1);
	for(int i = 1; i <= n; i++)
		sum[i] = sum[i - 1] + (s[i] == '1');
	for(int i = 1; i <= n; i++)
		rig[i] = i;
	for(int i = 1; i <= m; i++)
	{
		int l, r;
		cin >> l >> r;
		rig[l] = max(rig[l], r);
	}
	for(int i = 1; i <= n; i++)
		rig[i] = max(rig[i], rig[i - 1]);

	// for(int i = 1; i <= n; i++)
	// 	cerr << i << ' ' << rig[i] << endl;

	f[0][0] = 1;
	for(int i = 0; i <= n - 1; i++)
		for(int j = 0; j <= rig[i] - i + 1; j++)
			if(f[i][j])
			{
				int a = j + sum[rig[i + 1]] - sum[rig[i]];
				int b = rig[i + 1] - (i + 1) + 1 - a;
				if(a >= 1)
					add(f[i + 1][a - 1], f[i][j]);
				if(b >= 1)
					add(f[i + 1][a], f[i][j]);
			}
	int ans = f[n][0];
	cout << ans << endl;

	return 0;
}