#include<cstdio>
const int N=3005,mo=1e9+7;
int n,m,i,j,sum[N],r[N],f[N][N];char s[N];
void add(int &x,int k){x=(x+k)%mo;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);s[n+1]='0';
	for (i=1;i<=n+1;i++)
		r[i]=i,sum[i]=sum[i-1]+s[i]-48;
	for (;m--;){
		int x,y;scanf("%d%d",&x,&y);
		if (y>r[x]) r[x]=y;
	}
	for (i=2;i<=n+1;i++)
		if (r[i-1]>r[i]) r[i]=r[i-1];
	f[1][sum[r[1]]]=1;
	for (i=1;i<=n;i++) for (j=0;j<=n;j++) if (f[i][j]){
		int k=sum[r[i+1]]-sum[r[i]];
		if (r[i]-i+1-j) add(f[i+1][j+k],f[i][j]);
		if (j) add(f[i+1][j+k-1],f[i][j]);
	}
	printf("%d",f[n+1][0]);
}
### Prompt

Construct a CPP code solution to the problem outlined:

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,mo=1e9+7;
int n,m,i,j,sum[N],r[N],f[N][N];char s[N];
void add(int &x,int k){x=(x+k)%mo;}
int main(){
	scanf("%d%d%s",&n,&m,s+1);s[n+1]='0';
	for (i=1;i<=n+1;i++)
		r[i]=i,sum[i]=sum[i-1]+s[i]-48;
	for (;m--;){
		int x,y;scanf("%d%d",&x,&y);
		if (y>r[x]) r[x]=y;
	}
	for (i=2;i<=n+1;i++)
		if (r[i-1]>r[i]) r[i]=r[i-1];
	f[1][sum[r[1]]]=1;
	for (i=1;i<=n;i++) for (j=0;j<=n;j++) if (f[i][j]){
		int k=sum[r[i+1]]-sum[r[i]];
		if (r[i]-i+1-j) add(f[i+1][j+k],f[i][j]);
		if (j) add(f[i+1][j+k-1],f[i][j]);
	}
	printf("%d",f[n+1][0]);
}
```