#include<cstdio>
const int N=3005,md=1e9+7;
int f[N][N],rg[N],pre[N],n,m;
char ss[N];
inline void upd(int&a){a+=a>>31&md;}
int main(){
	scanf("%d%d%s",&n,&m,ss+1);
	for(int i=1;i<=n;++i)pre[i]=pre[i-1]+ss[i]-48,rg[i]=i;
	for(int i=1;i<=m;++i){
		int l,r;
		scanf("%d%d",&l,&r);
		if(rg[l]<r)rg[l]=r;
	}
	for(int i=2;i<=n+1;++i)if(rg[i-1]>rg[i])rg[i]=rg[i-1];
	f[1][pre[rg[1]]]=1;
	for(int i=1;i<=n;++i)
	for(int _1=0;_1<=n;++_1)
	if(int d=f[i][_1]){
		int s1=pre[rg[i+1]]-pre[rg[i]],_0=rg[i]-i+1-_1;
		if(_1)upd(f[i+1][_1-1+s1]+=d-md);if(_0)upd(f[i+1][_1+s1]+=d-md);
	}
	printf("%d\n",f[n+1][0]);
	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<cstdio>
const int N=3005,md=1e9+7;
int f[N][N],rg[N],pre[N],n,m;
char ss[N];
inline void upd(int&a){a+=a>>31&md;}
int main(){
	scanf("%d%d%s",&n,&m,ss+1);
	for(int i=1;i<=n;++i)pre[i]=pre[i-1]+ss[i]-48,rg[i]=i;
	for(int i=1;i<=m;++i){
		int l,r;
		scanf("%d%d",&l,&r);
		if(rg[l]<r)rg[l]=r;
	}
	for(int i=2;i<=n+1;++i)if(rg[i-1]>rg[i])rg[i]=rg[i-1];
	f[1][pre[rg[1]]]=1;
	for(int i=1;i<=n;++i)
	for(int _1=0;_1<=n;++_1)
	if(int d=f[i][_1]){
		int s1=pre[rg[i+1]]-pre[rg[i]],_0=rg[i]-i+1-_1;
		if(_1)upd(f[i+1][_1-1+s1]+=d-md);if(_0)upd(f[i+1][_1+s1]+=d-md);
	}
	printf("%d\n",f[n+1][0]);
	return 0;
}
```