#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 3005
#define mod 1000000007
using namespace std;
int n,m,ri[maxn],cnt[maxn],dp[maxn][maxn];
char s[maxn];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n+1;i++) ri[i]=i;
	for(int i=1;i<=n;i++) cnt[i]=cnt[i-1]+s[i]-'0';
	cnt[n+1]=cnt[n];
	while(m--)
	{
		int l,r;
		scanf("%d%d",&l,&r);
		ri[l]=max(ri[l],r);
	}
	for(int i=1;i<=n;i++) ri[i]=max(ri[i],ri[i-1]);
	dp[1][cnt[ri[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
			if(dp[i][j])
			{
				int l=ri[i]+1,r=ri[i+1];
				if(j) dp[i+1][j+cnt[r]-cnt[l-1]-1]=(dp[i+1][j+cnt[r]-cnt[l-1]-1]+dp[i][j])%mod;
				if(ri[i]-i+1-j) dp[i+1][j+cnt[r]-cnt[l-1]]=(dp[i+1][j+cnt[r]-cnt[l-1]]+dp[i][j])%mod;
			}
	printf("%d\n",(dp[n+1][0]+mod)%mod);
}
### 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<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 3005
#define mod 1000000007
using namespace std;
int n,m,ri[maxn],cnt[maxn],dp[maxn][maxn];
char s[maxn];
int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	for(int i=1;i<=n+1;i++) ri[i]=i;
	for(int i=1;i<=n;i++) cnt[i]=cnt[i-1]+s[i]-'0';
	cnt[n+1]=cnt[n];
	while(m--)
	{
		int l,r;
		scanf("%d%d",&l,&r);
		ri[l]=max(ri[l],r);
	}
	for(int i=1;i<=n;i++) ri[i]=max(ri[i],ri[i-1]);
	dp[1][cnt[ri[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
			if(dp[i][j])
			{
				int l=ri[i]+1,r=ri[i+1];
				if(j) dp[i+1][j+cnt[r]-cnt[l-1]-1]=(dp[i+1][j+cnt[r]-cnt[l-1]-1]+dp[i][j])%mod;
				if(ri[i]-i+1-j) dp[i+1][j+cnt[r]-cnt[l-1]]=(dp[i+1][j+cnt[r]-cnt[l-1]]+dp[i][j])%mod;
			}
	printf("%d\n",(dp[n+1][0]+mod)%mod);
}
```