#include<cstdio>
#include<algorithm>
using namespace std;
const int N=3030,P=1000000007;
int i,j,k,n,m,x,y;
int r[N],sum[N],f[N][N];
char s[N];
void Up(int &x,int y) {
	x=(x+y)%P;
}
int main() {
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (i=1;i<=n;i++) sum[i]=sum[i-1]+s[i]-'0';
	for (i=1;i<=n;i++) r[i]=i;
	for (i=1;i<=m;i++) {
		scanf("%d%d",&x,&y);
		r[x]=max(r[x],y);
	}
	f[1][sum[r[1]]]=1;
	int ma=r[1];
	for (i=1;i<n;i++) {
		int tmp=ma;
		ma=max(ma,r[i+1]);
		for (j=0;j<=tmp-i+1;j++) if (f[i][j]) {
			if (j) Up(f[i+1][j-1+sum[ma]-sum[tmp]],f[i][j]);
			if (j!=tmp-i+1) Up(f[i+1][j+sum[ma]-sum[tmp]],f[i][j]);
		}
	}
	printf("%d\n",(f[n][0]+f[n][1])%P);
}
### Prompt

Please create a solution in Cpp 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<cstdio>
#include<algorithm>
using namespace std;
const int N=3030,P=1000000007;
int i,j,k,n,m,x,y;
int r[N],sum[N],f[N][N];
char s[N];
void Up(int &x,int y) {
	x=(x+y)%P;
}
int main() {
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for (i=1;i<=n;i++) sum[i]=sum[i-1]+s[i]-'0';
	for (i=1;i<=n;i++) r[i]=i;
	for (i=1;i<=m;i++) {
		scanf("%d%d",&x,&y);
		r[x]=max(r[x],y);
	}
	f[1][sum[r[1]]]=1;
	int ma=r[1];
	for (i=1;i<n;i++) {
		int tmp=ma;
		ma=max(ma,r[i+1]);
		for (j=0;j<=tmp-i+1;j++) if (f[i][j]) {
			if (j) Up(f[i+1][j-1+sum[ma]-sum[tmp]],f[i][j]);
			if (j!=tmp-i+1) Up(f[i+1][j+sum[ma]-sum[tmp]],f[i][j]);
		}
	}
	printf("%d\n",(f[n][0]+f[n][1])%P);
}
```