#include<bits/stdc++.h>
using namespace std;

const int N = 3005, P = 1e9 + 7;
char s[N];
int n, m, d[N], f[N][N], num[N];

int main () {
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for (int i = 1; i <= n; ++i) d[i] = i, num[i] = num[i - 1] + (s[i] == '1');
	for (int i = 1, l, r; i <= m; ++i) {
		scanf("%d%d", &l, &r);
		if (r > d[l]) d[l] = r;
	}
	f[0][0] = 1;
	for (int i = 0, r = 0; i < n; ++i) {
		if (d[i + 1] > r) r = d[i + 1];
		for (int j = 0; j <= num[r]; ++j) {
			if (f[i][j] >= P) f[i][j] -= P;
			if (!f[i][j]) continue ;
			if (num[r] - j < r - i) f[i + 1][j] += f[i][j];
			if (num[r] - j > 0) f[i + 1][j + 1] += f[i][j];
		}
	}
	printf("%d\n", f[n][num[n]] % P);
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

const int N = 3005, P = 1e9 + 7;
char s[N];
int n, m, d[N], f[N][N], num[N];

int main () {
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for (int i = 1; i <= n; ++i) d[i] = i, num[i] = num[i - 1] + (s[i] == '1');
	for (int i = 1, l, r; i <= m; ++i) {
		scanf("%d%d", &l, &r);
		if (r > d[l]) d[l] = r;
	}
	f[0][0] = 1;
	for (int i = 0, r = 0; i < n; ++i) {
		if (d[i + 1] > r) r = d[i + 1];
		for (int j = 0; j <= num[r]; ++j) {
			if (f[i][j] >= P) f[i][j] -= P;
			if (!f[i][j]) continue ;
			if (num[r] - j < r - i) f[i + 1][j] += f[i][j];
			if (num[r] - j > 0) f[i + 1][j + 1] += f[i][j];
		}
	}
	printf("%d\n", f[n][num[n]] % P);
	return 0;
}
```