#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<vector>
#include<set>
using namespace std;
long long mod=1000000007;
char in[3100];
long long dp[3100][3100];
int r[3100];
int sum[3100];
int main(){
	int a,b;scanf("%d%d",&a,&b);
	scanf("%s",in);
	for(int i=0;i<a;i++)r[i]=i;
	for(int i=0;i<b;i++){
		int p,q;scanf("%d%d",&p,&q);
		p--;q--;
		r[p]=max(r[p],q);
	}
	for(int i=1;i<a;i++){
		r[i]=max(r[i],r[i-1]);
	}
	int cur=0;
	for(int i=0;i<a;i++){
		cur+=in[i]-'0';
		sum[i]=cur;
	}
	dp[0][0]=1;
	for(int i=0;i<a;i++){
		for(int j=0;j<=i;j++){
			if(!dp[i][j])continue;
			int t=sum[r[i]];
			if(j<=t&&t-j<=r[i]-i)dp[i+1][j]=(dp[i+1][j]+dp[i][j])%mod;
			if(j+1<=t&&t-j-1<=r[i]-i)dp[i+1][j+1]=(dp[i+1][j+1]+dp[i][j])%mod;
		}
	}
	printf("%lld\n",dp[a][sum[a-1]]);
}
### 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<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<string.h>
#include<vector>
#include<set>
using namespace std;
long long mod=1000000007;
char in[3100];
long long dp[3100][3100];
int r[3100];
int sum[3100];
int main(){
	int a,b;scanf("%d%d",&a,&b);
	scanf("%s",in);
	for(int i=0;i<a;i++)r[i]=i;
	for(int i=0;i<b;i++){
		int p,q;scanf("%d%d",&p,&q);
		p--;q--;
		r[p]=max(r[p],q);
	}
	for(int i=1;i<a;i++){
		r[i]=max(r[i],r[i-1]);
	}
	int cur=0;
	for(int i=0;i<a;i++){
		cur+=in[i]-'0';
		sum[i]=cur;
	}
	dp[0][0]=1;
	for(int i=0;i<a;i++){
		for(int j=0;j<=i;j++){
			if(!dp[i][j])continue;
			int t=sum[r[i]];
			if(j<=t&&t-j<=r[i]-i)dp[i+1][j]=(dp[i+1][j]+dp[i][j])%mod;
			if(j+1<=t&&t-j-1<=r[i]-i)dp[i+1][j+1]=(dp[i+1][j+1]+dp[i][j])%mod;
		}
	}
	printf("%lld\n",dp[a][sum[a-1]]);
}
```