#include <iostream>
#include <cstdio>
#include <cmath>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=3e3+5,Mod=1e9+7;
int f[N][N],n,m,a[N],r[N];
int Plus(int &x,int y){ return x=(x+y)%Mod; }
int main(){
	scanf("%d%d",&n,&m);
	rep(i,1,n){
		char c=getchar();
		while(c!='0'&&c!='1') c=getchar();
		a[i]=c-'0';
	}
	rep(i,1,n+1) a[i]=a[i]+a[i-1],r[i]=i;
	rep(i,1,m){
		int u,v;
		scanf("%d%d",&u,&v);
		r[u]=max(v,r[u]);
	}
	f[1][a[r[1]]]=1;
	rep(i,1,n) r[i]=max(r[i],r[i-1]);
	rep(i,1,n) rep(j,0,a[n]) if(f[i][j]){
		int L=r[i]+1,R=r[i+1],x=L-i-j,y=j;//x->0 y->1
		if(x) Plus(f[i+1][j+a[R]-a[L-1]],f[i][j]);
		if(y) Plus(f[i+1][j-1+a[R]-a[L-1]],f[i][j]);
	}
	printf("%d",f[n+1][0]%Mod);
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 <cmath>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
const int N=3e3+5,Mod=1e9+7;
int f[N][N],n,m,a[N],r[N];
int Plus(int &x,int y){ return x=(x+y)%Mod; }
int main(){
	scanf("%d%d",&n,&m);
	rep(i,1,n){
		char c=getchar();
		while(c!='0'&&c!='1') c=getchar();
		a[i]=c-'0';
	}
	rep(i,1,n+1) a[i]=a[i]+a[i-1],r[i]=i;
	rep(i,1,m){
		int u,v;
		scanf("%d%d",&u,&v);
		r[u]=max(v,r[u]);
	}
	f[1][a[r[1]]]=1;
	rep(i,1,n) r[i]=max(r[i],r[i-1]);
	rep(i,1,n) rep(j,0,a[n]) if(f[i][j]){
		int L=r[i]+1,R=r[i+1],x=L-i-j,y=j;//x->0 y->1
		if(x) Plus(f[i+1][j+a[R]-a[L-1]],f[i][j]);
		if(y) Plus(f[i+1][j-1+a[R]-a[L-1]],f[i][j]);
	}
	printf("%d",f[n+1][0]%Mod);
	return 0;
}
```