#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 3005;
const int mod = 1e9+7;

int n,m,a[maxn],dp[maxn][maxn];
char str[maxn];int s[maxn];
int main()
{
	#ifdef Amberframe
		freopen("arc065f.in","r",stdin);
		freopen("arc065f.out","w",stdout);
	#endif
	scanf("%d %d %s",&n,&m,str+1);
	for (int i=1;i<=n;i++)
		s[i]=s[i-1]+(str[i]=='1'),a[i]=i;
	for (int l,r,i=1;i<=m;i++)
		scanf("%d %d",&l,&r),a[l]=max(a[l],r);
	for (int i=1;i<=n;i++)
		a[i]=max(a[i-1],a[i]);
		
	dp[0][0]=1;
	for (int i=1;i<=n;i++)
	{
		int o=s[a[i]]-s[a[i-1]];
		for (int j=0;j<=s[a[i]];j++)
		if (dp[i-1][j]) {
			if (a[i]-i>=j+o) (dp[i][j+o]+=dp[i-1][j])%=mod;
			if (a[i]-i>=j+o-1) (dp[i][j+o-1]+=dp[i-1][j])%=mod;
		}
	}
	printf("%d",dp[n][0]);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <iostream>
#include <algorithm>
#include <memory.h>

using namespace std;
typedef long long LL;
const int maxn = 3005;
const int mod = 1e9+7;

int n,m,a[maxn],dp[maxn][maxn];
char str[maxn];int s[maxn];
int main()
{
	#ifdef Amberframe
		freopen("arc065f.in","r",stdin);
		freopen("arc065f.out","w",stdout);
	#endif
	scanf("%d %d %s",&n,&m,str+1);
	for (int i=1;i<=n;i++)
		s[i]=s[i-1]+(str[i]=='1'),a[i]=i;
	for (int l,r,i=1;i<=m;i++)
		scanf("%d %d",&l,&r),a[l]=max(a[l],r);
	for (int i=1;i<=n;i++)
		a[i]=max(a[i-1],a[i]);
		
	dp[0][0]=1;
	for (int i=1;i<=n;i++)
	{
		int o=s[a[i]]-s[a[i-1]];
		for (int j=0;j<=s[a[i]];j++)
		if (dp[i-1][j]) {
			if (a[i]-i>=j+o) (dp[i][j+o]+=dp[i-1][j])%=mod;
			if (a[i]-i>=j+o-1) (dp[i][j+o-1]+=dp[i-1][j])%=mod;
		}
	}
	printf("%d",dp[n][0]);
	return 0;
}
```