#include <bits/stdc++.h>
const int mu=1000000007;
int n,m,l,r,right[3005],sum[3005],f[3005][3005];
char s[3005];
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) right[i]=i;
	scanf("%s",s+1);
	for (int i=1;i<=n;i++) sum[i]=sum[i-1]+s[i]-'0';
	for (int i=1;i<=m;i++){
		scanf("%d%d",&l,&r);
		right[l]=std::max(right[l],r);
	}
	for (int i=1;i<=n;i++) right[i]=std::max(right[i],right[i-1]);
	f[0][0]=1;
	for (int i=1;i<=n;i++){
		int t=sum[right[i]]-sum[right[i-1]];
		for (int j=0;j<=right[i];j++){
			if (!f[i-1][j]) continue;
			if (j+t>0) f[i][j+t-1]=(f[i][j+t-1]+f[i-1][j])%mu;
			if (right[i]-i+1>j+t) f[i][j+t]=(f[i][j+t]+f[i-1][j])%mu; 
		}
		//if (i==right[i]) for (int j=1;j<=n;j++) f[i][j]=0;
	}
	printf("%d\n",f[n][0]);
}
### Prompt

Your challenge is to write 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>
const int mu=1000000007;
int n,m,l,r,right[3005],sum[3005],f[3005][3005];
char s[3005];
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) right[i]=i;
	scanf("%s",s+1);
	for (int i=1;i<=n;i++) sum[i]=sum[i-1]+s[i]-'0';
	for (int i=1;i<=m;i++){
		scanf("%d%d",&l,&r);
		right[l]=std::max(right[l],r);
	}
	for (int i=1;i<=n;i++) right[i]=std::max(right[i],right[i-1]);
	f[0][0]=1;
	for (int i=1;i<=n;i++){
		int t=sum[right[i]]-sum[right[i-1]];
		for (int j=0;j<=right[i];j++){
			if (!f[i-1][j]) continue;
			if (j+t>0) f[i][j+t-1]=(f[i][j+t-1]+f[i-1][j])%mu;
			if (right[i]-i+1>j+t) f[i][j+t]=(f[i][j+t]+f[i-1][j])%mu; 
		}
		//if (i==right[i]) for (int j=1;j<=n;j++) f[i][j]=0;
	}
	printf("%d\n",f[n][0]);
}
```