#include<stdio.h>
#include<algorithm>
using namespace std;
int D[3010][3010], P[3010], Mod = 1000000007;
int S[3010], n, m;
char p[3010];
int main(){
    int i, a, b, j;
    scanf("%d%d",&n,&m);
    scanf("%s",p+1);
    for(i=1;i<=n;i++){
        P[i]=i;
        S[i]=S[i-1]+p[i]-'0';
    }
    for(i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        for(j=a;j<=b;j++)P[j]=max(P[j],b);
    }
    D[0][0]=1;
    for(i=0;i<n;i++){
        int bb, ee;
        bb = max(0, i - (P[i] - S[P[i]])), ee = min(i, S[P[i]]);
        for(j=bb;j<=ee;j++){
            D[i+1][j]=(D[i+1][j]+D[i][j])%Mod;
            D[i+1][j+1]=(D[i+1][j+1]+D[i][j])%Mod;
        }
    }
    printf("%d\n",D[n][S[n]]);
}
### Prompt

In CPP, your task is to solve 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<stdio.h>
#include<algorithm>
using namespace std;
int D[3010][3010], P[3010], Mod = 1000000007;
int S[3010], n, m;
char p[3010];
int main(){
    int i, a, b, j;
    scanf("%d%d",&n,&m);
    scanf("%s",p+1);
    for(i=1;i<=n;i++){
        P[i]=i;
        S[i]=S[i-1]+p[i]-'0';
    }
    for(i=1;i<=m;i++){
        scanf("%d%d",&a,&b);
        for(j=a;j<=b;j++)P[j]=max(P[j],b);
    }
    D[0][0]=1;
    for(i=0;i<n;i++){
        int bb, ee;
        bb = max(0, i - (P[i] - S[P[i]])), ee = min(i, S[P[i]]);
        for(j=bb;j<=ee;j++){
            D[i+1][j]=(D[i+1][j]+D[i][j])%Mod;
            D[i+1][j+1]=(D[i+1][j+1]+D[i][j])%Mod;
        }
    }
    printf("%d\n",D[n][S[n]]);
}
```