#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
const int maxn = 3005;
const int mo = 1e9 + 7;
using namespace std;
int f[2][maxn], s[maxn], Max[maxn];
int n, m;
int main(){
	scanf("%d%d", &n, &m); getchar();
	for (int i = 1; i <= n; i++) s[i] = s[i - 1] + getchar() - '0';
	for (int l, r, i = 1; i <= m; i++){
		scanf("%d%d", &l, &r);
		Max[l] = max(Max[l], r);
	}
	for (int i = 1; i <= n; i++) Max[i] = max(i, max(Max[i], Max[i - 1]));
	f[0][0] = 1; int T = 1;
	for (int i = 1; i <= n; T ^= 1, i++){
		memset(f[T], 0, sizeof f[T]);
		for (int j = 0; j <= n; j++){
			if (j - 1 + s[Max[i]] - s[Max[i - 1]] >= 0) (f[T][j - 1 + s[Max[i]] - s[Max[i - 1]]] += f[T ^ 1][j]) %= mo;
			if (j + s[Max[i]] - s[Max[i - 1]] < Max[i] - (i - 1)) (f[T][j + s[Max[i]] - s[Max[i - 1]]] += f[T ^ 1][j]) %= mo;
		}
	}
	printf("%d\n", f[T ^ 1][0]);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
const int maxn = 3005;
const int mo = 1e9 + 7;
using namespace std;
int f[2][maxn], s[maxn], Max[maxn];
int n, m;
int main(){
	scanf("%d%d", &n, &m); getchar();
	for (int i = 1; i <= n; i++) s[i] = s[i - 1] + getchar() - '0';
	for (int l, r, i = 1; i <= m; i++){
		scanf("%d%d", &l, &r);
		Max[l] = max(Max[l], r);
	}
	for (int i = 1; i <= n; i++) Max[i] = max(i, max(Max[i], Max[i - 1]));
	f[0][0] = 1; int T = 1;
	for (int i = 1; i <= n; T ^= 1, i++){
		memset(f[T], 0, sizeof f[T]);
		for (int j = 0; j <= n; j++){
			if (j - 1 + s[Max[i]] - s[Max[i - 1]] >= 0) (f[T][j - 1 + s[Max[i]] - s[Max[i - 1]]] += f[T ^ 1][j]) %= mo;
			if (j + s[Max[i]] - s[Max[i - 1]] < Max[i] - (i - 1)) (f[T][j + s[Max[i]] - s[Max[i - 1]]] += f[T ^ 1][j]) %= mo;
		}
	}
	printf("%d\n", f[T ^ 1][0]);
	return 0;
}
```