#include<bits/stdc++.h>
#define N (3009)
#define MOD (1000000007)
using namespace std;

int n,m,x,y,R[N],sum[N],f[N][N];
char s[N];

int main()
{
	cin>>n>>m>>s+1; s[n+1]='0';
	for (int i=1; i<=n+1; ++i)
		R[i]=i, sum[i]=sum[i-1]+s[i]-'0';
	for (int i=1; i<=m; ++i)
		cin>>x>>y, R[x]=max(R[x],y);
	for (int i=2; i<=n+1; ++i)
		R[i]=max(R[i],R[i-1]);
	f[1][sum[R[1]]]=1;
	for (int i=1; i<=n; ++i)
		for (int j=0; j<=n; ++j)
		{
			int k=sum[R[i+1]]-sum[R[i]];
			if (j) (f[i+1][j+k-1]+=f[i][j])%=MOD;
			if (R[i]-i>=j) (f[i+1][j+k]+=f[i][j])%=MOD;
		}
	cout<<f[n+1][0]<<endl;
}
### Prompt

In Cpp, your task is to solve 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>
#define N (3009)
#define MOD (1000000007)
using namespace std;

int n,m,x,y,R[N],sum[N],f[N][N];
char s[N];

int main()
{
	cin>>n>>m>>s+1; s[n+1]='0';
	for (int i=1; i<=n+1; ++i)
		R[i]=i, sum[i]=sum[i-1]+s[i]-'0';
	for (int i=1; i<=m; ++i)
		cin>>x>>y, R[x]=max(R[x],y);
	for (int i=2; i<=n+1; ++i)
		R[i]=max(R[i],R[i-1]);
	f[1][sum[R[1]]]=1;
	for (int i=1; i<=n; ++i)
		for (int j=0; j<=n; ++j)
		{
			int k=sum[R[i+1]]-sum[R[i]];
			if (j) (f[i+1][j+k-1]+=f[i][j])%=MOD;
			if (R[i]-i>=j) (f[i+1][j+k]+=f[i][j])%=MOD;
		}
	cout<<f[n+1][0]<<endl;
}
```