#include <bits/stdc++.h>
using namespace std;

const int mo=int(1e9+7);
int n,m,r[3010],sum[3010],f[3010][3010];
char s[3010];

void work()
{
	scanf("%d %d\n%s",&n,&m,s+1);
	for (int i=1; i<=n; i++)  sum[i]=sum[i-1]+(s[i]=='1'),r[i]=i;
	for (int i=1,x,y; i<=m; i++)  scanf("%d %d",&x,&y),r[x]=max(r[x],y);
	for (int i=1; i<=n; i++)  r[i]=max(r[i],r[i-1]);
	f[0][0]=1;
	for (int i=1; i<=n; i++)
		for (int j=0; j<=sum[r[i-1]]; j++)
			if (f[i-1][j])
				{
					if (sum[r[i]]>j)  f[i][j+1]=(f[i][j+1]+f[i-1][j])%mo;
					if (sum[r[i]]-j<r[i]-i+1)  f[i][j]=(f[i][j]+f[i-1][j])%mo;
				}
	printf("%d",f[n][sum[n]]);
}

int main()
{
	work();
	return 0;
}

### Prompt

Please create a solution in cpp 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 int mo=int(1e9+7);
int n,m,r[3010],sum[3010],f[3010][3010];
char s[3010];

void work()
{
	scanf("%d %d\n%s",&n,&m,s+1);
	for (int i=1; i<=n; i++)  sum[i]=sum[i-1]+(s[i]=='1'),r[i]=i;
	for (int i=1,x,y; i<=m; i++)  scanf("%d %d",&x,&y),r[x]=max(r[x],y);
	for (int i=1; i<=n; i++)  r[i]=max(r[i],r[i-1]);
	f[0][0]=1;
	for (int i=1; i<=n; i++)
		for (int j=0; j<=sum[r[i-1]]; j++)
			if (f[i-1][j])
				{
					if (sum[r[i]]>j)  f[i][j+1]=(f[i][j+1]+f[i-1][j])%mo;
					if (sum[r[i]]-j<r[i]-i+1)  f[i][j]=(f[i][j]+f[i-1][j])%mo;
				}
	printf("%d",f[n][sum[n]]);
}

int main()
{
	work();
	return 0;
}

```