#include<bits/stdc++.h>
using namespace std;
using ll=long long;
static const ll mod=1000000007;
using P=pair<ll,ll>;
ll sum1[3005];ll sum2[3005];
ll N,M;
string s;
ll dp[3005][3005];
int main(){
  cin>>N>>M>>s;
  vector<ll>A(N);
  for(ll i=0;i<N;i++)
    A[i]=s[i]-'0';
  vector<P>C(M);
  for(int i=0;i<M;i++){
  ll l,r;cin>>l>>r;P p(l,r);
    C[i]=p;
  }vector<ll>A1(N);vector<ll>A2(N);
  for(ll i=0;i<N;i++){A1[i]=A[i];A2[i]=A[i];}
  for(ll i=0;i<M;i++){
    ll l=C[i].first;ll r=C[i].second;
  sort(A1.begin()+l-1,A1.begin()+r);
     sort(A2.begin()+l-1,A2.begin()+r);
    reverse(A2.begin()+l-1,A2.begin()+r);
    }for(ll i=1;i<=N;i++){
  sum1[i]=sum1[i-1]+A1[i-1];
    sum2[i]=sum2[i-1]+A2[i-1];
  }dp[0][0]=1;
  for(ll i=1;i<=N;i++)
    for(ll j=sum1[i];j<=sum2[i];j++){
      dp[i][j]=dp[i-1][j];
      if(0<=j-1)
        dp[i][j]=(dp[i][j]+dp[i-1][j-1])%mod;
    }cout<<dp[N][sum1[N]]<<endl;
  return 0;
}
### 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<bits/stdc++.h>
using namespace std;
using ll=long long;
static const ll mod=1000000007;
using P=pair<ll,ll>;
ll sum1[3005];ll sum2[3005];
ll N,M;
string s;
ll dp[3005][3005];
int main(){
  cin>>N>>M>>s;
  vector<ll>A(N);
  for(ll i=0;i<N;i++)
    A[i]=s[i]-'0';
  vector<P>C(M);
  for(int i=0;i<M;i++){
  ll l,r;cin>>l>>r;P p(l,r);
    C[i]=p;
  }vector<ll>A1(N);vector<ll>A2(N);
  for(ll i=0;i<N;i++){A1[i]=A[i];A2[i]=A[i];}
  for(ll i=0;i<M;i++){
    ll l=C[i].first;ll r=C[i].second;
  sort(A1.begin()+l-1,A1.begin()+r);
     sort(A2.begin()+l-1,A2.begin()+r);
    reverse(A2.begin()+l-1,A2.begin()+r);
    }for(ll i=1;i<=N;i++){
  sum1[i]=sum1[i-1]+A1[i-1];
    sum2[i]=sum2[i-1]+A2[i-1];
  }dp[0][0]=1;
  for(ll i=1;i<=N;i++)
    for(ll j=sum1[i];j<=sum2[i];j++){
      dp[i][j]=dp[i-1][j];
      if(0<=j-1)
        dp[i][j]=(dp[i][j]+dp[i-1][j-1])%mod;
    }cout<<dp[N][sum1[N]]<<endl;
  return 0;
}
```