#include <cstdio> 
#include <iostream>
using namespace std;
const int N=3010,mo=1e9+7;
int n,m,f[N][N],mx[N],sum[N];
int main()
{
    scanf("%d%d\n",&n,&m);
    for (int i=1;i<=n;i++) sum[i]=getchar()-'0';
    for (int i=1;i<=n+1;i++) sum[i]+=sum[i-1];
    for (int i=1;i<=n+1;i++) mx[i]=i;
    for (int i=1,x,y;i<=m;i++) scanf("%d%d",&x,&y),mx[x]=max(mx[x],y);
    for (int i=1;i<=n;i++) mx[i]=max(mx[i],mx[i-1]) ;
    f[1][sum[mx[1]]]=1;
    for (int i=1,l,r,x,y;i<=n;i++) for (int j=0;j<=n;j++) if (f[i][j])
	{
    	l=mx[i]+1,r=mx[i+1],x=j,y=mx[i]-i+1-j;
        if (y) (f[i+1][j+sum[r]-sum[l-1]]+=f[i][j])%=mo;
        if (x) (f[i+1][j-1+sum[r]-sum[l-1]]+=f[i][j])%=mo;
    }
    printf("%d\n",f[n+1][0]);
}
### Prompt

Please formulate 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 <cstdio> 
#include <iostream>
using namespace std;
const int N=3010,mo=1e9+7;
int n,m,f[N][N],mx[N],sum[N];
int main()
{
    scanf("%d%d\n",&n,&m);
    for (int i=1;i<=n;i++) sum[i]=getchar()-'0';
    for (int i=1;i<=n+1;i++) sum[i]+=sum[i-1];
    for (int i=1;i<=n+1;i++) mx[i]=i;
    for (int i=1,x,y;i<=m;i++) scanf("%d%d",&x,&y),mx[x]=max(mx[x],y);
    for (int i=1;i<=n;i++) mx[i]=max(mx[i],mx[i-1]) ;
    f[1][sum[mx[1]]]=1;
    for (int i=1,l,r,x,y;i<=n;i++) for (int j=0;j<=n;j++) if (f[i][j])
	{
    	l=mx[i]+1,r=mx[i+1],x=j,y=mx[i]-i+1-j;
        if (y) (f[i+1][j+sum[r]-sum[l-1]]+=f[i][j])%=mo;
        if (x) (f[i+1][j-1+sum[r]-sum[l-1]]+=f[i][j])%=mo;
    }
    printf("%d\n",f[n+1][0]);
}
```