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

const int mod = 1e9+7;

long long int dp[3001][3002];
int sum[3001];
char s[3001];
int r[3001];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>s[i];
		sum[i]=s[i]-'0';
		sum[i]+=sum[i-1];
	}
	for(int i=1;i<=n;i++)
		r[i]=i;
	for(int i=1,x,y;i<=m;i++){
		cin>>x>>y;
		r[x]=max(r[x],y);
	}
	dp[0][0]=1;
	int now=0;
	for(int i=0;i<n;i++){
		int tnow=max(now,r[i+1]);
		for(int j=0;j<=3000;j++){
			int ones,zeros;
			if(r[i+1]>now)
				ones=j+sum[tnow]-sum[now];
			else
				ones=j;
			zeros=tnow-ones-i;
			if(ones)
				dp[i+1][ones-1]=(dp[i+1][ones-1]+dp[i][j])%mod;
			if(zeros)
				dp[i+1][ones]=(dp[i+1][ones]+dp[i][j])%mod;
		}
		now=tnow;	
	}
	cout<<dp[n][0]<<endl;
}

### Prompt

Generate 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;

const int mod = 1e9+7;

long long int dp[3001][3002];
int sum[3001];
char s[3001];
int r[3001];
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n,m;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>s[i];
		sum[i]=s[i]-'0';
		sum[i]+=sum[i-1];
	}
	for(int i=1;i<=n;i++)
		r[i]=i;
	for(int i=1,x,y;i<=m;i++){
		cin>>x>>y;
		r[x]=max(r[x],y);
	}
	dp[0][0]=1;
	int now=0;
	for(int i=0;i<n;i++){
		int tnow=max(now,r[i+1]);
		for(int j=0;j<=3000;j++){
			int ones,zeros;
			if(r[i+1]>now)
				ones=j+sum[tnow]-sum[now];
			else
				ones=j;
			zeros=tnow-ones-i;
			if(ones)
				dp[i+1][ones-1]=(dp[i+1][ones-1]+dp[i][j])%mod;
			if(zeros)
				dp[i+1][ones]=(dp[i+1][ones]+dp[i][j])%mod;
		}
		now=tnow;	
	}
	cout<<dp[n][0]<<endl;
}

```