#include<bits/stdc++.h>
const int N=3010;
const int yql=1e9+7;
using namespace std;
int a[N],dp[N][N],n,m,s[N];
char p[N];
inline int read(){
	int f=1,x=0;char ch;
	do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
	do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
	return f*x;
}
int main(){
	n=read();m=read();
	scanf("%s",p+1);
	for(int i=1;i<=n;i++){a[i]=i;s[i]=s[i-1];if(p[i]=='1')s[i]++;}
	//for(int i=1;i<=n;i++)printf("%d%c",s[i],i==n?'\n':' ');
	for(int i=1;i<=m;i++){
		int l=read(),r=read();
		for(int j=l;j<=r;j++)a[j]=max(a[j],r);
	}
	dp[0][0]=1;
	for(int i=1;i<=n;i++){
		for(int j=0;j<=i;j++){
			if(j>s[a[i]])continue;
			if(i-j>a[i]-s[a[i]])continue;
			dp[i][j]=dp[i-1][j];
			if(j)dp[i][j]=(dp[i][j]+dp[i-1][j-1])%yql;
		}
	}
	printf("%d\n",dp[n][s[n]]);
}
### Prompt

Your challenge is to write 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>
const int N=3010;
const int yql=1e9+7;
using namespace std;
int a[N],dp[N][N],n,m,s[N];
char p[N];
inline int read(){
	int f=1,x=0;char ch;
	do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');
	do{x=x*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');
	return f*x;
}
int main(){
	n=read();m=read();
	scanf("%s",p+1);
	for(int i=1;i<=n;i++){a[i]=i;s[i]=s[i-1];if(p[i]=='1')s[i]++;}
	//for(int i=1;i<=n;i++)printf("%d%c",s[i],i==n?'\n':' ');
	for(int i=1;i<=m;i++){
		int l=read(),r=read();
		for(int j=l;j<=r;j++)a[j]=max(a[j],r);
	}
	dp[0][0]=1;
	for(int i=1;i<=n;i++){
		for(int j=0;j<=i;j++){
			if(j>s[a[i]])continue;
			if(i-j>a[i]-s[a[i]])continue;
			dp[i][j]=dp[i-1][j];
			if(j)dp[i][j]=(dp[i][j]+dp[i-1][j-1])%yql;
		}
	}
	printf("%d\n",dp[n][s[n]]);
}
```