#include <cstdio>

const int N = 3e3 + 7, MOD = 1e9 + 7;
int n, m, total, corder[N], prev[N], dp[N][N];
char now;

inline int max(int a, int b){return a > b?a:b;}

int main(){
	scanf("%d %d", &n, &m);
	for (register int i = 1;i <= n;i ++){
		while (now = getchar(), now != '1' && now != '0')continue;
		prev[i] = prev[i - 1], corder[i] = i;
		if (now == '1')total ++, prev[i] ++;
	}
	prev[n + 1] = prev[n], corder[n + 1] = n + 1;
	for (register int i = 1, x, y;i <= m;i ++){
		scanf("%d %d", &x, &y);
		corder[x] = max(corder[x], y);
	}
	for (register int i = 1;i <= n + 1;i ++)corder[i] = max(corder[i], corder[i - 1]);
	dp[1][prev[corder[1]]] = 1;
	for (register int i = 1;i <= n;i ++)
		for (register int j = 0;j <= total;j ++)
			if (dp[i][j]){
				if (j)dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]] - 1] = (dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]] - 1] + dp[i][j])%MOD;
				if (corder[i] + 1 - i - j)dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]]] = (dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]]] + dp[i][j])%MOD;
			}
	return !printf("%d", dp[n + 1][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 <cstdio>

const int N = 3e3 + 7, MOD = 1e9 + 7;
int n, m, total, corder[N], prev[N], dp[N][N];
char now;

inline int max(int a, int b){return a > b?a:b;}

int main(){
	scanf("%d %d", &n, &m);
	for (register int i = 1;i <= n;i ++){
		while (now = getchar(), now != '1' && now != '0')continue;
		prev[i] = prev[i - 1], corder[i] = i;
		if (now == '1')total ++, prev[i] ++;
	}
	prev[n + 1] = prev[n], corder[n + 1] = n + 1;
	for (register int i = 1, x, y;i <= m;i ++){
		scanf("%d %d", &x, &y);
		corder[x] = max(corder[x], y);
	}
	for (register int i = 1;i <= n + 1;i ++)corder[i] = max(corder[i], corder[i - 1]);
	dp[1][prev[corder[1]]] = 1;
	for (register int i = 1;i <= n;i ++)
		for (register int j = 0;j <= total;j ++)
			if (dp[i][j]){
				if (j)dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]] - 1] = (dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]] - 1] + dp[i][j])%MOD;
				if (corder[i] + 1 - i - j)dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]]] = (dp[i + 1][j + prev[corder[i + 1]] - prev[corder[i]]] + dp[i][j])%MOD;
			}
	return !printf("%d", dp[n + 1][0]);
}
```