#include <bits/stdc++.h>
using namespace std;
#define ll int64_t

const ll mod=1e9+7;

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);
	int n,m;
	cin>>n>>m;
	vector<char> s(n+5),t(n+5);
	vector<vector<ll>> dp(n+5,vector<ll>(n+5));
	for(int i=1;i<=n;++i){
		cin>>s[i];
		t[i]=s[i];
	}
	for(int i=0;i<m;++i){
		int l,r;
		cin>>l>>r;
		sort(s.begin()+l,s.begin()+r+1);
		sort(t.begin()+l,t.begin()+r+1,greater<char>());
	}
	dp[0][0]=1;
	int lo=0,hi=0;
	for(int i=1;i<=n;++i){
		lo+=s[i]=='1';
		hi+=t[i]=='1';
		for(int j=lo;j<=hi;++j)
			(dp[i][j]=dp[i-1][j]+(j>0?dp[i-1][j-1]:0))%=mod;
	}
	cout<<dp[n][lo]<<'\n';
}


### 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>
using namespace std;
#define ll int64_t

const ll mod=1e9+7;

int main() {
	ios_base::sync_with_stdio(0);cin.tie(0);
	int n,m;
	cin>>n>>m;
	vector<char> s(n+5),t(n+5);
	vector<vector<ll>> dp(n+5,vector<ll>(n+5));
	for(int i=1;i<=n;++i){
		cin>>s[i];
		t[i]=s[i];
	}
	for(int i=0;i<m;++i){
		int l,r;
		cin>>l>>r;
		sort(s.begin()+l,s.begin()+r+1);
		sort(t.begin()+l,t.begin()+r+1,greater<char>());
	}
	dp[0][0]=1;
	int lo=0,hi=0;
	for(int i=1;i<=n;++i){
		lo+=s[i]=='1';
		hi+=t[i]=='1';
		for(int j=lo;j<=hi;++j)
			(dp[i][j]=dp[i-1][j]+(j>0?dp[i-1][j-1]:0))%=mod;
	}
	cout<<dp[n][lo]<<'\n';
}


```