#include<bits/stdc++.h>
#define N 3005
using namespace std;
const int mo=1e9+7;
int n,m,len,a[N],l[N],r[N],L[N],R[N],f[N],g[N],C[N][N],vis[N],ans;
int main()
{
	int i,j,k;
	scanf("%d %d\n",&n,&m);
	for(i=1;i<=n;i++)
		a[i]=a[i-1]+getchar()-'0';
	for(i=0;i<=n;i++){
		C[i][0]=C[i][i]=1;
		for(j=1;j<i;j++)
			C[i][j]=(C[i-1][j-1]+C[i-1][j])%mo;
	  }
	f[0]=1;
	for(i=1;i<=m;i++)
		scanf("%d %d",&L[i],&R[i]);
	for(i=1;i<=m;i++)
		for(j=1;j<=m;j++)
			if(L[i]>=L[j]&&R[i]<=R[j]){
				if(L[i]!=L[j]||R[i]!=R[j]) vis[i]=1;
				else vis[i]=(j<i);
				break;
			  }
	for(i=1;i<=m;i++)
		if(!vis[i])
			++len,l[len]=L[i],r[len]=R[i];
	l[0]=1,m=len;
	for(i=0;i<m;i++){
		for(j=0;j<=n;j++) g[j]=f[j],f[j]=0;
		for(j=0;j<=r[i]-l[i]+1;j++)
			if(r[i]<l[i+1])
				f[a[r[i+1]]-a[l[i+1]-1]]=(f[a[r[i+1]]-a[l[i+1]-1]]+1LL*g[j]*C[r[i]-l[i]+1][j])%mo;
			else{
				for(k=max(j-r[i]+l[i+1]-1,0);k<=j&&k<=l[i+1]-l[i];k++)
					f[j+a[r[i+1]]-a[r[i]]-k]=(f[j+a[r[i+1]]-a[r[i]]-k]+1LL*g[j]*C[l[i+1]-l[i]][k])%mo;
			  }
	  }
	for(i=0;i<=r[m]-l[m]+1;i++)
		ans=(ans+1LL*f[i]*C[r[m]-l[m]+1][i])%mo;
	cout<<ans;
	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<bits/stdc++.h>
#define N 3005
using namespace std;
const int mo=1e9+7;
int n,m,len,a[N],l[N],r[N],L[N],R[N],f[N],g[N],C[N][N],vis[N],ans;
int main()
{
	int i,j,k;
	scanf("%d %d\n",&n,&m);
	for(i=1;i<=n;i++)
		a[i]=a[i-1]+getchar()-'0';
	for(i=0;i<=n;i++){
		C[i][0]=C[i][i]=1;
		for(j=1;j<i;j++)
			C[i][j]=(C[i-1][j-1]+C[i-1][j])%mo;
	  }
	f[0]=1;
	for(i=1;i<=m;i++)
		scanf("%d %d",&L[i],&R[i]);
	for(i=1;i<=m;i++)
		for(j=1;j<=m;j++)
			if(L[i]>=L[j]&&R[i]<=R[j]){
				if(L[i]!=L[j]||R[i]!=R[j]) vis[i]=1;
				else vis[i]=(j<i);
				break;
			  }
	for(i=1;i<=m;i++)
		if(!vis[i])
			++len,l[len]=L[i],r[len]=R[i];
	l[0]=1,m=len;
	for(i=0;i<m;i++){
		for(j=0;j<=n;j++) g[j]=f[j],f[j]=0;
		for(j=0;j<=r[i]-l[i]+1;j++)
			if(r[i]<l[i+1])
				f[a[r[i+1]]-a[l[i+1]-1]]=(f[a[r[i+1]]-a[l[i+1]-1]]+1LL*g[j]*C[r[i]-l[i]+1][j])%mo;
			else{
				for(k=max(j-r[i]+l[i+1]-1,0);k<=j&&k<=l[i+1]-l[i];k++)
					f[j+a[r[i+1]]-a[r[i]]-k]=(f[j+a[r[i+1]]-a[r[i]]-k]+1LL*g[j]*C[l[i+1]-l[i]][k])%mo;
			  }
	  }
	for(i=0;i<=r[m]-l[m]+1;i++)
		ans=(ans+1LL*f[i]*C[r[m]-l[m]+1][i])%mo;
	cout<<ans;
	return 0;
}

```