#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int p=1e9+7;
int n,m,i,j,x,y,re[5010],f[3010][3010],a[3010][2];
char s[3010];
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(i=1;i<=n;i++){
		a[i][0]=a[i-1][0];a[i][1]=a[i-1][1];
		if(s[i]=='0')a[i][0]++;
		 else a[i][1]++;
		re[i]=i;
	}
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		re[x]=max(re[x],y);
	}
	for(i=1;i<=n;i++)re[i]=max(re[i],re[i-1]);
	f[0][0]=1;
	for(i=1;i<=n;i++)
	 for(j=0;j<i;j++)if(f[i-1][j]){
	 	if(j+1<=a[re[i]][0])f[i][j+1]=(f[i][j+1]+f[i-1][j])%p;
	 	if(i-j<=a[re[i]][1])f[i][j]=(f[i][j]+f[i-1][j])%p;
	 }
	printf("%d",f[n][a[n][0]]);
}
### 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<bits/stdc++.h>
#define ll long long
using namespace std;
const int p=1e9+7;
int n,m,i,j,x,y,re[5010],f[3010][3010],a[3010][2];
char s[3010];
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(i=1;i<=n;i++){
		a[i][0]=a[i-1][0];a[i][1]=a[i-1][1];
		if(s[i]=='0')a[i][0]++;
		 else a[i][1]++;
		re[i]=i;
	}
	for(i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		re[x]=max(re[x],y);
	}
	for(i=1;i<=n;i++)re[i]=max(re[i],re[i-1]);
	f[0][0]=1;
	for(i=1;i<=n;i++)
	 for(j=0;j<i;j++)if(f[i-1][j]){
	 	if(j+1<=a[re[i]][0])f[i][j+1]=(f[i][j+1]+f[i-1][j])%p;
	 	if(i-j<=a[re[i]][1])f[i][j]=(f[i][j]+f[i-1][j])%p;
	 }
	printf("%d",f[n][a[n][0]]);
}
```