#include<bits/stdc++.h>
using namespace std;
const long long md=1e9+7;
long long n,m,cc[3005],a[3005],dp[3005][3005];
char s[3005];
int main()
{
	scanf("%lld%lld%s",&n,&m,s+1);
	for(long long i=1;i<=n;a[i]=i,i++)cc[i]=cc[i-1]+(s[i]-'0');
	for(long long i=1,l,r;i<=m;i++)
	{
		scanf("%lld%lld",&l,&r);
		for(long long j=l;j<=r;j++)a[j]=max(a[j],r);
	}
	dp[0][0]=1;
	for(long long i=1;i<=n;i++)
	{
		for(long long j=0;j<=i;j++)
		{
			if(j>cc[a[i]])continue;
			if(i-j>a[i]-cc[a[i]])continue;
			dp[i][j]=dp[i-1][j];
			if(j)dp[i][j]=(dp[i][j]+dp[i-1][j-1])%md;
		}
	}
	printf("%lld\n",dp[n][cc[n]]);
	return 0;
}
### Prompt

Your task is to create 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 long long md=1e9+7;
long long n,m,cc[3005],a[3005],dp[3005][3005];
char s[3005];
int main()
{
	scanf("%lld%lld%s",&n,&m,s+1);
	for(long long i=1;i<=n;a[i]=i,i++)cc[i]=cc[i-1]+(s[i]-'0');
	for(long long i=1,l,r;i<=m;i++)
	{
		scanf("%lld%lld",&l,&r);
		for(long long j=l;j<=r;j++)a[j]=max(a[j],r);
	}
	dp[0][0]=1;
	for(long long i=1;i<=n;i++)
	{
		for(long long j=0;j<=i;j++)
		{
			if(j>cc[a[i]])continue;
			if(i-j>a[i]-cc[a[i]])continue;
			dp[i][j]=dp[i-1][j];
			if(j)dp[i][j]=(dp[i][j]+dp[i-1][j-1])%md;
		}
	}
	printf("%lld\n",dp[n][cc[n]]);
	return 0;
}
```