#include <bits/stdc++.h>
using namespace std;

const long long mod=1000000007ll;
void ad(long long &a,long long b){a+=b;a%=mod;}
int main()
{
  int n,m;
  scanf("%d%d",&n,&m);
  char s[3010];
  scanf("%s",s);
  int R[3010];
  for(int i=0;i<n;i++){
    R[i]=i;
  }
  for(int i=0;i<m;i++){
    int l,r;
    scanf("%d%d",&l,&r);
    l--;
    r--;
    R[l]=max(R[l],r);
  }
  int S[3010];
  S[0]=0;
  for(int i=0;i<n;i++){
    S[i+1]=S[i]+(s[i]=='1'?1:0);
  }
  long long dp[2][3010]={0};
  bool b=0;
  dp[b][0]=1ll;
  int r=0;
  for(int i=0;i<n;i++){
    for(int j=0;j<=n;j++){
      dp[!b][j]=0ll;
    }
    int nr=max(R[i]+1,r);
    int p=S[nr]-S[r];
    r=nr;
    //printf("%d %d %d\n",i,r,p);
    for(int j=0;j<=n;j++){
      if(dp[b][j]==0){
	continue;
      }
      if(r-i-(j+p)>0){
	ad(dp[!b][j+p],dp[b][j]);
      }
      if(j+p>0){
	ad(dp[!b][j+p-1],dp[b][j]);
      }
    }
    b=!b;
  }
  printf("%lld\n",dp[b][0]);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;

const long long mod=1000000007ll;
void ad(long long &a,long long b){a+=b;a%=mod;}
int main()
{
  int n,m;
  scanf("%d%d",&n,&m);
  char s[3010];
  scanf("%s",s);
  int R[3010];
  for(int i=0;i<n;i++){
    R[i]=i;
  }
  for(int i=0;i<m;i++){
    int l,r;
    scanf("%d%d",&l,&r);
    l--;
    r--;
    R[l]=max(R[l],r);
  }
  int S[3010];
  S[0]=0;
  for(int i=0;i<n;i++){
    S[i+1]=S[i]+(s[i]=='1'?1:0);
  }
  long long dp[2][3010]={0};
  bool b=0;
  dp[b][0]=1ll;
  int r=0;
  for(int i=0;i<n;i++){
    for(int j=0;j<=n;j++){
      dp[!b][j]=0ll;
    }
    int nr=max(R[i]+1,r);
    int p=S[nr]-S[r];
    r=nr;
    //printf("%d %d %d\n",i,r,p);
    for(int j=0;j<=n;j++){
      if(dp[b][j]==0){
	continue;
      }
      if(r-i-(j+p)>0){
	ad(dp[!b][j+p],dp[b][j]);
      }
      if(j+p>0){
	ad(dp[!b][j+p-1],dp[b][j]);
      }
    }
    b=!b;
  }
  printf("%lld\n",dp[b][0]);
  return 0;
}

```