#include<bits/stdc++.h>
typedef long long i64;
const int P=1e9+7,N=3007;
char s0[N];
int n,m,rs[N],p=0,f[N],ans=1;
inline int fix(int x){return x-=P,x+(x>>31&P);}
int main(){
	scanf("%d%d%s",&n,&m,s0+1);
	for(int i=1;i<=n;++i)rs[i]=i;
	for(int i=1,x,y;i<=m;++i){
		scanf("%d%d",&x,&y);
		if(y>rs[x])rs[x]=y;
	}
	f[0]=1;
	for(int i=1,mr=0;i<=n;++i){
		if(rs[i]>mr){
			int c1=0;
			for(int j=mr+1;j<=rs[i];++j)c1+=s0[j]-'0';
			if(c1)for(int j=mr-i+1;j>=0;--j)f[j+c1]=f[j],f[j]=0;
			mr=rs[i];
		}
		for(int j=0;j<=mr-i;++j)f[j]=fix(f[j]+f[j+1]);
		f[mr-i+1]=0;
	}
	printf("%d\n",f[0]);
	return 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>
typedef long long i64;
const int P=1e9+7,N=3007;
char s0[N];
int n,m,rs[N],p=0,f[N],ans=1;
inline int fix(int x){return x-=P,x+(x>>31&P);}
int main(){
	scanf("%d%d%s",&n,&m,s0+1);
	for(int i=1;i<=n;++i)rs[i]=i;
	for(int i=1,x,y;i<=m;++i){
		scanf("%d%d",&x,&y);
		if(y>rs[x])rs[x]=y;
	}
	f[0]=1;
	for(int i=1,mr=0;i<=n;++i){
		if(rs[i]>mr){
			int c1=0;
			for(int j=mr+1;j<=rs[i];++j)c1+=s0[j]-'0';
			if(c1)for(int j=mr-i+1;j>=0;--j)f[j+c1]=f[j],f[j]=0;
			mr=rs[i];
		}
		for(int j=0;j<=mr-i;++j)f[j]=fix(f[j]+f[j+1]);
		f[mr-i+1]=0;
	}
	printf("%d\n",f[0]);
	return 0;
}
```