#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
#define N 3005

const LL mod=1000000007;
LL ans,n,m,C[N][N],f[N],g[N],a[N];
char s[N];

void upd(LL &x,LL y){x=(x+y)%mod;}

int main(){
	scanf("%lld%lld%s",&n,&m,s+1);
	for (LL i=0;i<=n;++i){
		C[i][0]=1;
		for (LL j=1;j<=i;++j) C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
	}
	for (LL i=1;i<=n;++i) a[i]=s[i]-'0';
	f[0]=1;
	ans=1;
	LL l=1,r=0;
	for (LL i=1;i<=m;++i){
		LL x,y; scanf("%lld%lld",&x,&y);
		if (y<=r) continue;
		if (x>r){
			LL sum=0;
			for (LL j=0;j<=r-l+1;++j){
				upd(sum,f[j]*C[r-l+1][j]);
			}
			ans=ans*sum%mod;
			memset(f,0,sizeof f);
			f[0]=1;
			l=x; r=x-1;
		}
		LL tmp=0;
		for (LL j=r+1;j<=y;++j){
			tmp+=a[j];
		}
		LL len=x-l;
		memset(g,0,sizeof g);
		for (LL j=0;j<=n;++j){
			for (LL k=0;k<=len&&k<=j;++k){
				if (j-k<=r-l+1-len) upd(g[j-k+tmp],f[j]*C[len][k]);
			}
		}
		memcpy(f,g,sizeof f);
		l=x; r=y;
	}
	LL sum=0;
	for (LL j=0;j<=r-l+1;++j){
		upd(sum,f[j]*C[r-l+1][j]);
	}
	ans=ans*sum%mod;
	printf("%lld\n",ans);
	
	return 0;
}
			
### Prompt

Please formulate a cpp solution 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;

typedef long long LL;
#define N 3005

const LL mod=1000000007;
LL ans,n,m,C[N][N],f[N],g[N],a[N];
char s[N];

void upd(LL &x,LL y){x=(x+y)%mod;}

int main(){
	scanf("%lld%lld%s",&n,&m,s+1);
	for (LL i=0;i<=n;++i){
		C[i][0]=1;
		for (LL j=1;j<=i;++j) C[i][j]=(C[i-1][j-1]+C[i-1][j])%mod;
	}
	for (LL i=1;i<=n;++i) a[i]=s[i]-'0';
	f[0]=1;
	ans=1;
	LL l=1,r=0;
	for (LL i=1;i<=m;++i){
		LL x,y; scanf("%lld%lld",&x,&y);
		if (y<=r) continue;
		if (x>r){
			LL sum=0;
			for (LL j=0;j<=r-l+1;++j){
				upd(sum,f[j]*C[r-l+1][j]);
			}
			ans=ans*sum%mod;
			memset(f,0,sizeof f);
			f[0]=1;
			l=x; r=x-1;
		}
		LL tmp=0;
		for (LL j=r+1;j<=y;++j){
			tmp+=a[j];
		}
		LL len=x-l;
		memset(g,0,sizeof g);
		for (LL j=0;j<=n;++j){
			for (LL k=0;k<=len&&k<=j;++k){
				if (j-k<=r-l+1-len) upd(g[j-k+tmp],f[j]*C[len][k]);
			}
		}
		memcpy(f,g,sizeof f);
		l=x; r=y;
	}
	LL sum=0;
	for (LL j=0;j<=r-l+1;++j){
		upd(sum,f[j]*C[r-l+1][j]);
	}
	ans=ans*sum%mod;
	printf("%lld\n",ans);
	
	return 0;
}
			
```