#include<bits/stdc++.h>

const int N = 3010;
const int moder = 1e9 + 7;

int dp[N][N], a[N], pre[N], r[N], n, m;
char s[N];

int main(){
	scanf("%d%d%s", &n, &m, s);
	for (int i = 1; i <= n; ++ i){
		r[i] = i;
	}
	for (int i = 0, l, right; i < m; ++ i){
		scanf("%d%d", &l, &right);
		for (int j = l; j <= right; ++ j){
			r[j] = std::max(r[j], right);
		}
	}
	for (int i = 1; i <= n; ++ i){
		a[i] = s[i - 1] - '0';
		pre[i] = pre[i - 1] + (a[i] == 1);
	}
	dp[0][0] = 1;
	for (int i = 1; i <= n + 1; ++ i){
		for (int j = 0; j <= i - 1; ++ j){
			int restone = pre[r[i]] - j;
			if (restone < 0){
				continue;
			}
			if (restone > 0){
				dp[i][j + 1] = (dp[i][j + 1] + dp[i - 1][j]) % moder;
			}
			if (restone != r[i] - i + 1){
				dp[i][j] = (dp[i][j] + dp[i - 1][j]) % moder;
			}
		}
	}
	printf("%d\n", dp[n][pre[n]]);
	return 0;
}
### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include<bits/stdc++.h>

const int N = 3010;
const int moder = 1e9 + 7;

int dp[N][N], a[N], pre[N], r[N], n, m;
char s[N];

int main(){
	scanf("%d%d%s", &n, &m, s);
	for (int i = 1; i <= n; ++ i){
		r[i] = i;
	}
	for (int i = 0, l, right; i < m; ++ i){
		scanf("%d%d", &l, &right);
		for (int j = l; j <= right; ++ j){
			r[j] = std::max(r[j], right);
		}
	}
	for (int i = 1; i <= n; ++ i){
		a[i] = s[i - 1] - '0';
		pre[i] = pre[i - 1] + (a[i] == 1);
	}
	dp[0][0] = 1;
	for (int i = 1; i <= n + 1; ++ i){
		for (int j = 0; j <= i - 1; ++ j){
			int restone = pre[r[i]] - j;
			if (restone < 0){
				continue;
			}
			if (restone > 0){
				dp[i][j + 1] = (dp[i][j + 1] + dp[i - 1][j]) % moder;
			}
			if (restone != r[i] - i + 1){
				dp[i][j] = (dp[i][j] + dp[i - 1][j]) % moder;
			}
		}
	}
	printf("%d\n", dp[n][pre[n]]);
	return 0;
}
```