#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1000000007;
int dp[3030][3030];
int r[3030],cnt[3030];
char s[3030];
int main()
{
	int n,m,l1,r1;
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	memset(cnt,0,sizeof(cnt));
	for(int i=1;i<=n;i++)
		cnt[i]=s[i]-'0';
	for(int i=1;i<=n;i++)
		cnt[i]+=cnt[i-1];
	for(int i=1;i<=n;i++)
		r[i]=i;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&l1,&r1);
		r[l1]=max(r[l1],r1);
	}
	for(int i=1;i<=n;i++)
		r[i]=max(r[i],r[i-1]);
	memset(dp,0,sizeof(dp));
	dp[1][0]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<i;j++)
			if(dp[i][j])
			{
				if(cnt[r[i]]!=j)
					(dp[i+1][j+1]+=dp[i][j])%=mod;
				if(i-1-j!=r[i]-cnt[r[i]])
					(dp[i+1][j]+=dp[i][j])%=mod;
			}
	printf("%d",dp[n+1][cnt[n]]);
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1000000007;
int dp[3030][3030];
int r[3030],cnt[3030];
char s[3030];
int main()
{
	int n,m,l1,r1;
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	memset(cnt,0,sizeof(cnt));
	for(int i=1;i<=n;i++)
		cnt[i]=s[i]-'0';
	for(int i=1;i<=n;i++)
		cnt[i]+=cnt[i-1];
	for(int i=1;i<=n;i++)
		r[i]=i;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&l1,&r1);
		r[l1]=max(r[l1],r1);
	}
	for(int i=1;i<=n;i++)
		r[i]=max(r[i],r[i-1]);
	memset(dp,0,sizeof(dp));
	dp[1][0]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<i;j++)
			if(dp[i][j])
			{
				if(cnt[r[i]]!=j)
					(dp[i+1][j+1]+=dp[i][j])%=mod;
				if(i-1-j!=r[i]-cnt[r[i]])
					(dp[i+1][j]+=dp[i][j])%=mod;
			}
	printf("%d",dp[n+1][cnt[n]]);
	return 0;
}
```