#include<cstdio>
const int mod=1000000007;
int N,M,R[3010],sum[3010],f[3010][3010];
char S[3010];
int main(){
	scanf("%d%d%s",&N,&M,S);
	for(int i=0;i<N;i++)R[i]=i+1;
	for(int i=0;i<M;i++){
		int l,r;
		scanf("%d%d",&l,&r);
		if(r>R[l-1])R[l-1]=r;
	}
	for(int i=0;i<N;i++)if(R[i]>R[i+1])R[i+1]=R[i];
	for(int i=0;i<N;i++)sum[i+1]=sum[i]+S[i]-'0';
	f[0][0]=1;
	for(int i=0;i<N;i++)for(int j=0;j<=sum[N];j++)if(f[i][j]){
		if(sum[R[i]]-j<R[i]-i)(f[i+1][j]+=f[i][j])%=mod;
		if(sum[R[i]]-j>0)(f[i+1][j+1]+=f[i][j])%=mod;
	}
	printf("%d\n",f[N][sum[N]]);
}
### Prompt

Your task is to create 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<cstdio>
const int mod=1000000007;
int N,M,R[3010],sum[3010],f[3010][3010];
char S[3010];
int main(){
	scanf("%d%d%s",&N,&M,S);
	for(int i=0;i<N;i++)R[i]=i+1;
	for(int i=0;i<M;i++){
		int l,r;
		scanf("%d%d",&l,&r);
		if(r>R[l-1])R[l-1]=r;
	}
	for(int i=0;i<N;i++)if(R[i]>R[i+1])R[i+1]=R[i];
	for(int i=0;i<N;i++)sum[i+1]=sum[i]+S[i]-'0';
	f[0][0]=1;
	for(int i=0;i<N;i++)for(int j=0;j<=sum[N];j++)if(f[i][j]){
		if(sum[R[i]]-j<R[i]-i)(f[i+1][j]+=f[i][j])%=mod;
		if(sum[R[i]]-j>0)(f[i+1][j+1]+=f[i][j])%=mod;
	}
	printf("%d\n",f[N][sum[N]]);
}
```