#include<bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin() , (x).end()
#define rall(x) (x).rbegin() , (x).rend()
#define len(x) (int)(x).size()

typedef long long ll;
typedef long double ld;

using namespace std;

const int MOD = (int)1e9 + 7;
const int N = 3001;
int dp[N][N];
int sum(int x , int y){
    x += y; if(x >= MOD)x-=MOD;
    return x;
}
signed main(){
    ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);
    int n , m;cin >> n >> m;
    string s;cin >> s;
    vector < int > r(n);
    iota(all(r) , 0);
    for(int i = 0 ; i < m; ++i){
        int l , _r;cin >> l >> _r;
        --l , --_r;
        r[l] = max(r[l] , _r);
    }
    dp[0][0] = 1;
    int all = 0 , j = 0;
    for(int i = 0; i < n; ++i){
        while(j <= r[i])
            all += s[j] - '0' , j++;
        for(int was = 0 ;was <= i; ++was){
            /// put 1
            if(was + 1 <= all)
                dp[i + 1][was + 1] = sum(dp[i+1][was+1],dp[i][was]);
            /// put 0
            if((j - all) - (i - was))
                dp[i + 1][was] = sum(dp[i+1][was] , dp[i][was]);
        }
    }
    int answer = 0;
    for(int i = 0; i <= n; ++i)
        answer = sum(answer , dp[n][i]);
    cout << answer;
    return 0;
}

### Prompt

Please formulate 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>
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin() , (x).end()
#define rall(x) (x).rbegin() , (x).rend()
#define len(x) (int)(x).size()

typedef long long ll;
typedef long double ld;

using namespace std;

const int MOD = (int)1e9 + 7;
const int N = 3001;
int dp[N][N];
int sum(int x , int y){
    x += y; if(x >= MOD)x-=MOD;
    return x;
}
signed main(){
    ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);
    int n , m;cin >> n >> m;
    string s;cin >> s;
    vector < int > r(n);
    iota(all(r) , 0);
    for(int i = 0 ; i < m; ++i){
        int l , _r;cin >> l >> _r;
        --l , --_r;
        r[l] = max(r[l] , _r);
    }
    dp[0][0] = 1;
    int all = 0 , j = 0;
    for(int i = 0; i < n; ++i){
        while(j <= r[i])
            all += s[j] - '0' , j++;
        for(int was = 0 ;was <= i; ++was){
            /// put 1
            if(was + 1 <= all)
                dp[i + 1][was + 1] = sum(dp[i+1][was+1],dp[i][was]);
            /// put 0
            if((j - all) - (i - was))
                dp[i + 1][was] = sum(dp[i+1][was] , dp[i][was]);
        }
    }
    int answer = 0;
    for(int i = 0; i <= n; ++i)
        answer = sum(answer , dp[n][i]);
    cout << answer;
    return 0;
}

```