#include<cstdio>
#include<algorithm>
#define MAXN 3006
#define MOD 1000000007
using namespace std;
int n,m,ll,rr,r[MAXN],d[MAXN][MAXN],sum[MAXN];
char s[MAXN];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n+1;i++)
	{
		sum[i]=sum[i-1];
		if(i<=n) sum[i]+=s[i]-'0';
		r[i]=i;
	}
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&ll,&rr);
		r[ll]=max(r[ll],rr);
	}
	for(int i=1;i<=n;i++)
		r[i]=max(r[i],r[i-1]);
	d[1][sum[r[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
		{
			if(!d[i][j]) continue;
			if(j) d[i+1][j-1+sum[r[i+1]]-sum[r[i]]]=(d[i+1][j-1+sum[r[i+1]]-sum[r[i]]]+d[i][j])%MOD;
			if(r[i]-(i-1)-j) d[i+1][j+sum[r[i+1]]-sum[r[i]]]=(d[i+1][j+sum[r[i+1]]-sum[r[i]]]+d[i][j])%MOD;
		}
	printf("%d",(d[n+1][0]+MOD)%MOD);
}
### Prompt

Please provide a CPP coded solution 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<cstdio>
#include<algorithm>
#define MAXN 3006
#define MOD 1000000007
using namespace std;
int n,m,ll,rr,r[MAXN],d[MAXN][MAXN],sum[MAXN];
char s[MAXN];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n+1;i++)
	{
		sum[i]=sum[i-1];
		if(i<=n) sum[i]+=s[i]-'0';
		r[i]=i;
	}
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&ll,&rr);
		r[ll]=max(r[ll],rr);
	}
	for(int i=1;i<=n;i++)
		r[i]=max(r[i],r[i-1]);
	d[1][sum[r[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
		{
			if(!d[i][j]) continue;
			if(j) d[i+1][j-1+sum[r[i+1]]-sum[r[i]]]=(d[i+1][j-1+sum[r[i+1]]-sum[r[i]]]+d[i][j])%MOD;
			if(r[i]-(i-1)-j) d[i+1][j+sum[r[i+1]]-sum[r[i]]]=(d[i+1][j+sum[r[i+1]]-sum[r[i]]]+d[i][j])%MOD;
		}
	printf("%d",(d[n+1][0]+MOD)%MOD);
}
```