#include <bits/stdc++.h>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

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

void add(int &x, int y){
	x += y;
	if(x >= Mod) x -= Mod;
}

int dp[N][N];
int n, m;
int L[N], sum[N], R[N];
char S[N];

int main(){
	scanf("%d%d", &n, &m);
	scanf("%s", S + 1);
	For(i, 1, n) sum[i] = sum[i - 1] + (S[i] == '1');
	For(i, 1, m) scanf("%d%d", &L[i], &R[i]);

	dp[0][0] = 1;
	int p = 1, mx = 0;
	For(i, 1, n){
		mx = max(mx, i);
		while(p <= m && L[p] == i){
			mx = max(mx, R[p]);
			++p;
		}

		For(j, 0, i - 1){
			int v = dp[i - 1][j];
			if(!v) continue;
			if(j + 1 <= sum[mx]) add(dp[i][j + 1], v);
			if(i - j <= mx - sum[mx]) add(dp[i][j], v);
		}
	}
	printf("%d\n", dp[n][sum[n]]);
	return 0;
}
### Prompt

Please formulate a cpp solution to 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>

#define For(i, j, k) for(int i = j; i <= k; i++)

using namespace std;

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

void add(int &x, int y){
	x += y;
	if(x >= Mod) x -= Mod;
}

int dp[N][N];
int n, m;
int L[N], sum[N], R[N];
char S[N];

int main(){
	scanf("%d%d", &n, &m);
	scanf("%s", S + 1);
	For(i, 1, n) sum[i] = sum[i - 1] + (S[i] == '1');
	For(i, 1, m) scanf("%d%d", &L[i], &R[i]);

	dp[0][0] = 1;
	int p = 1, mx = 0;
	For(i, 1, n){
		mx = max(mx, i);
		while(p <= m && L[p] == i){
			mx = max(mx, R[p]);
			++p;
		}

		For(j, 0, i - 1){
			int v = dp[i - 1][j];
			if(!v) continue;
			if(j + 1 <= sum[mx]) add(dp[i][j + 1], v);
			if(i - j <= mx - sum[mx]) add(dp[i][j], v);
		}
	}
	printf("%d\n", dp[n][sum[n]]);
	return 0;
}
```