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

#define int long long
#define fi first
#define se second
#define pb push_back
#define ins insert
#define er erase

typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<ii, ii> iiii;

const int oo = 1e18 + 7, mod = 1e9 + 7;

const int N = 3005;

int n, m, r[N], dp[N][N], pref[N];
string s;

signed main(){
	ios_base::sync_with_stdio(0);
	cin >> n >> m;
	cin >> s;
    s = '*' + s;
    for(int i = 1; i <= n; i++) pref[i] = (pref[i - 1] + (s[i] - '0'));
	for(int i = 1; i <= m; i++){
	    int le, ri;
	    cin >> le >> ri;
	    r[le] = max(r[le], ri);
	}
	for(int i = 2; i <= n; i++) if(i <= r[i - 1]) r[i] = max(r[i], r[i - 1]);
	dp[0][0] = 1;
	for(int i = 1; i <= n; i++){
	    for(int j = 0; j <= i; j++){
	        if(!r[i]){
	            if(j >= (s[i] - '0')) dp[i][j] = dp[i - 1][j - (s[i] - '0')];
	            continue;
	        }
	        else{
	            if(r[i] - pref[r[i]] < (i - j)) continue;
	            if(pref[r[i]] < j) continue;
	            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1]) % mod;
	        }
	    }
	}
	cout << dp[n][pref[n]];
}


### 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;

#define int long long
#define fi first
#define se second
#define pb push_back
#define ins insert
#define er erase

typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<ii, ii> iiii;

const int oo = 1e18 + 7, mod = 1e9 + 7;

const int N = 3005;

int n, m, r[N], dp[N][N], pref[N];
string s;

signed main(){
	ios_base::sync_with_stdio(0);
	cin >> n >> m;
	cin >> s;
    s = '*' + s;
    for(int i = 1; i <= n; i++) pref[i] = (pref[i - 1] + (s[i] - '0'));
	for(int i = 1; i <= m; i++){
	    int le, ri;
	    cin >> le >> ri;
	    r[le] = max(r[le], ri);
	}
	for(int i = 2; i <= n; i++) if(i <= r[i - 1]) r[i] = max(r[i], r[i - 1]);
	dp[0][0] = 1;
	for(int i = 1; i <= n; i++){
	    for(int j = 0; j <= i; j++){
	        if(!r[i]){
	            if(j >= (s[i] - '0')) dp[i][j] = dp[i - 1][j - (s[i] - '0')];
	            continue;
	        }
	        else{
	            if(r[i] - pref[r[i]] < (i - j)) continue;
	            if(pref[r[i]] < j) continue;
	            dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1]) % mod;
	        }
	    }
	}
	cout << dp[n][pref[n]];
}


```