#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;

ll add(ll a,ll b){
  return (a+b)%mod;
}

#define MAX 3005
int n,m;
string s;
int a[MAX],b[MAX];
int x[MAX];

ll dp[MAX][MAX];

int main(){
  cin>>n>>m>>s;
  for(int i=0;i<n;i++){
    x[i]=i;
    a[i]=(s[i]=='0'?1:0) + (i?a[i-1]:0);
    b[i]=(s[i]=='1'?1:0) + (i?b[i-1]:0);
  }
  for(int i=0;i<m;i++){
    int l,r;
    cin>>l>>r;
    l--;
    r--;
    for(int j=l;j<r;j++){
      x[j]=max(x[j],r);
    }
  }

  dp[0][0]=1;
  for(int i=0;i<n;i++){
    int p=x[i];
    for(int j=0;j<=i;j++){
      if(dp[i][j]==0)continue;
      int A=j;
      int B=i-A;
      if( A+1 <= a[p] ){
        dp[i+1][A+1]=add(dp[i+1][A+1],dp[i][j]);
      }
      if( B+1 <= b[p] ){
        dp[i+1][A]=add(dp[i+1][A],dp[i][j]);
      }
    }
  }

  cout<<dp[n][ a[n-1] ]<<endl;
  return 0;
}

### Prompt

Your challenge is to write 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9+7;

ll add(ll a,ll b){
  return (a+b)%mod;
}

#define MAX 3005
int n,m;
string s;
int a[MAX],b[MAX];
int x[MAX];

ll dp[MAX][MAX];

int main(){
  cin>>n>>m>>s;
  for(int i=0;i<n;i++){
    x[i]=i;
    a[i]=(s[i]=='0'?1:0) + (i?a[i-1]:0);
    b[i]=(s[i]=='1'?1:0) + (i?b[i-1]:0);
  }
  for(int i=0;i<m;i++){
    int l,r;
    cin>>l>>r;
    l--;
    r--;
    for(int j=l;j<r;j++){
      x[j]=max(x[j],r);
    }
  }

  dp[0][0]=1;
  for(int i=0;i<n;i++){
    int p=x[i];
    for(int j=0;j<=i;j++){
      if(dp[i][j]==0)continue;
      int A=j;
      int B=i-A;
      if( A+1 <= a[p] ){
        dp[i+1][A+1]=add(dp[i+1][A+1],dp[i][j]);
      }
      if( B+1 <= b[p] ){
        dp[i+1][A]=add(dp[i+1][A],dp[i][j]);
      }
    }
  }

  cout<<dp[n][ a[n-1] ]<<endl;
  return 0;
}

```