#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int mod=1e9+7;
int f[3005][3005],sum[3005],n,m,mx[3005];
char s[3005];
int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) sum[i]=sum[i-1]+(s[i]=='1');
	for(int i=1,l,r;i<=m;i++){
		scanf("%d%d",&l,&r);
		for(int p=l;p<=r;p++) mx[p]=max(mx[p],r);
	}
	f[0][0]=1;
	for(int i=1;i<=n;i++){
		if(!mx[i]){
			for(int p=0;p<=i;p++){
				if(s[i]=='1'&&p) f[i][p]=f[i-1][p-1];
				if(s[i]=='0') f[i][p]=f[i-1][p];
			}
			continue;
		}
		for(int p=0;p<=i;p++){
			if(p) f[i][p]=add(f[i-1][p],f[i-1][p-1]);
			else f[i][p]=f[i-1][p];
			if(sum[mx[i]]<p||(mx[i]-sum[mx[i]])<i-p) f[i][p]=0;
		}
	}
	printf("%d",f[n][sum[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 <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int mod=1e9+7;
int f[3005][3005],sum[3005],n,m,mx[3005];
char s[3005];
int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
int main(){
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++) sum[i]=sum[i-1]+(s[i]=='1');
	for(int i=1,l,r;i<=m;i++){
		scanf("%d%d",&l,&r);
		for(int p=l;p<=r;p++) mx[p]=max(mx[p],r);
	}
	f[0][0]=1;
	for(int i=1;i<=n;i++){
		if(!mx[i]){
			for(int p=0;p<=i;p++){
				if(s[i]=='1'&&p) f[i][p]=f[i-1][p-1];
				if(s[i]=='0') f[i][p]=f[i-1][p];
			}
			continue;
		}
		for(int p=0;p<=i;p++){
			if(p) f[i][p]=add(f[i-1][p],f[i-1][p-1]);
			else f[i][p]=f[i-1][p];
			if(sum[mx[i]]<p||(mx[i]-sum[mx[i]])<i-p) f[i][p]=0;
		}
	}
	printf("%d",f[n][sum[n]]);
	return 0;
}
```