#include<cstdio>
#include<algorithm>
#define int long long
using std::max;
const int mod=1000000007;
int n,m,s[3010],right[3010];
int x,y,f[3010][3010];
char S[3010];
signed main(){
	scanf("%lld%lld",&n,&m);
	scanf("%s",S+1);
	for(int i=1;i<=n;i++)
		s[i]=S[i]-'0';
	for(int i=1;i<=n+1;i++)
		s[i]+=s[i-1];
	for(int i=1;i<=n+1;i++)
		right[i]=i;
	for(int i=1;i<=m;i++){
		scanf("%lld%lld",&x,&y);
		right[x]=max(right[x],y);
	}
	for(int i=1;i<=n;i++)
		right[i]=max(right[i],right[i-1]);
	f[1][s[right[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
			if(f[i][j]){
				int l=right[i]+1,r=right[i+1];
				int bo1=j,bo2=right[i]-i+1-j;
				if(bo2)(f[i+1][j+s[r]-s[l-1]]+=f[i][j])%=mod;
				if(bo1)(f[i+1][j-1+s[r]-s[l-1]]+=f[i][j])%=mod;
			}
	printf("%lld\n",f[n+1][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<cstdio>
#include<algorithm>
#define int long long
using std::max;
const int mod=1000000007;
int n,m,s[3010],right[3010];
int x,y,f[3010][3010];
char S[3010];
signed main(){
	scanf("%lld%lld",&n,&m);
	scanf("%s",S+1);
	for(int i=1;i<=n;i++)
		s[i]=S[i]-'0';
	for(int i=1;i<=n+1;i++)
		s[i]+=s[i-1];
	for(int i=1;i<=n+1;i++)
		right[i]=i;
	for(int i=1;i<=m;i++){
		scanf("%lld%lld",&x,&y);
		right[x]=max(right[x],y);
	}
	for(int i=1;i<=n;i++)
		right[i]=max(right[i],right[i-1]);
	f[1][s[right[1]]]=1;
	for(int i=1;i<=n;i++)
		for(int j=0;j<=n;j++)
			if(f[i][j]){
				int l=right[i]+1,r=right[i+1];
				int bo1=j,bo2=right[i]-i+1-j;
				if(bo2)(f[i+1][j+s[r]-s[l-1]]+=f[i][j])%=mod;
				if(bo1)(f[i+1][j-1+s[r]-s[l-1]]+=f[i][j])%=mod;
			}
	printf("%lld\n",f[n+1][0]);
}
```