#include<bits/stdc++.h>
#define N 3005
using namespace std;
const int p = 1000000007;
int sum[N],a[N],n,m;
char s[N];
int r[N],f[N][N];
int main()
{
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    for(int i=1;i<=n;i++)a[i]=s[i]-'0',sum[i]=sum[i-1]+a[i];
    int t1,t2;
    for(int i=1;i<=n;i++)r[i]=i;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&t1,&t2);
        r[t1]=max(r[t1],t2);
    }
    f[0][0]=1;int now=0;
    for(int i=1;i<=n;i++)
    {
        now=max(now,r[i]);
        for(int j=0;j<=n;j++)
        {
            if(f[i-1][j])
            {
                if(sum[now]!=j+now-i+1)(f[i][j]+=f[i-1][j])%=p;
                if(sum[now]!=j)(f[i][j+1]+=f[i-1][j])%=p;
            }
        }
    }
    int ans=0;
    for(int i=0;i<=n;i++)ans+=f[n][i],ans%=p;
    cout<<ans<<endl;
    return 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<bits/stdc++.h>
#define N 3005
using namespace std;
const int p = 1000000007;
int sum[N],a[N],n,m;
char s[N];
int r[N],f[N][N];
int main()
{
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    for(int i=1;i<=n;i++)a[i]=s[i]-'0',sum[i]=sum[i-1]+a[i];
    int t1,t2;
    for(int i=1;i<=n;i++)r[i]=i;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&t1,&t2);
        r[t1]=max(r[t1],t2);
    }
    f[0][0]=1;int now=0;
    for(int i=1;i<=n;i++)
    {
        now=max(now,r[i]);
        for(int j=0;j<=n;j++)
        {
            if(f[i-1][j])
            {
                if(sum[now]!=j+now-i+1)(f[i][j]+=f[i-1][j])%=p;
                if(sum[now]!=j)(f[i][j+1]+=f[i-1][j])%=p;
            }
        }
    }
    int ans=0;
    for(int i=0;i<=n;i++)ans+=f[n][i],ans%=p;
    cout<<ans<<endl;
    return 0;
}

```