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

Solution (in CPP):

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

#define int long long
typedef vector<int>vint;
typedef pair<int,int>pint;
typedef vector<pint>vpint;
#define rep(i,n) for(int i=0;i<(n);i++)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define all(v) (v).begin(),(v).end()
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define pb push_back
#define fi first
#define se second
template<typename A,typename B>inline void chmin(A &a,B b){if(a>b)a=b;}
template<typename A,typename B>inline void chmax(A &a,B b){if(a<b)a=b;}

int N,M;

string S;
const int mod=1000000007;
int to[3333];
int sum[3333];

int dp[3333][3333];

signed main(){
    cin>>N>>M;
    cin>>S;
    rep(i,N)sum[i+1]=sum[i]+(S[i]=='1');
    rep(i,N)to[i+1]=i+1;
    rep(i,M){
        int l,r;
        cin>>l>>r;
        chmax(to[l],r);
    }
    for(int i=1;i<N;i++)chmax(to[i],to[i-1]);
    dp[0][0]=1;
    for(int i=0;i<N;i++){
        for(int j=max(i-(to[i]-sum[to[i]]),0ll);j<=min(i,sum[to[i]]);j++){
            if(to[i+1]-sum[to[i+1]]>=i-j+1)dp[i+1][j]=(dp[i+1][j]+dp[i][j])%mod;
            if(sum[to[i+1]]>=j+1)dp[i+1][j+1]=(dp[i+1][j+1]+dp[i][j])%mod;
        }
    }
    cout<<dp[N][sum[N]]<<endl;
    return 0;
}