#include<bits/stdc++.h>
#include<cstring>
#define N 3005
using namespace std;
const int mod=1e9+7;
int n,m;
char t[N];
int s[N],val[N],dp[2][N];
int main(){
	//freopen("in.txt","r",stdin);
	scanf("%d%d%s",&n,&m,t+1);
	for(int i=1;i<=n;i++) s[i]=t[i]-'0'+s[i-1];
	int cnt=0;
	for(int i=1,ml,mr;i<=m;i++)
		scanf("%d%d",&ml,&mr),val[ml]=max(val[ml],mr);
	for(int i=1;i<=n;i++) val[i]=max(max(i,val[i]),val[i-1]); 
	int now=0;
	dp[now][0]=1;
	for(int i=1;i<=n;i++,now^=1){
		memset(dp[now^1],0,sizeof(dp[0]));
		for(int j=max(0,s[val[i]]-(val[i]-i));j<=s[val[i]];j++)
			dp[now^1][j]=(dp[now][j]+ (j==0?0:dp[now][j-1]) )%mod;
	}
	cout<<dp[now][s[n]]<<endl;
	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<bits/stdc++.h>
#include<cstring>
#define N 3005
using namespace std;
const int mod=1e9+7;
int n,m;
char t[N];
int s[N],val[N],dp[2][N];
int main(){
	//freopen("in.txt","r",stdin);
	scanf("%d%d%s",&n,&m,t+1);
	for(int i=1;i<=n;i++) s[i]=t[i]-'0'+s[i-1];
	int cnt=0;
	for(int i=1,ml,mr;i<=m;i++)
		scanf("%d%d",&ml,&mr),val[ml]=max(val[ml],mr);
	for(int i=1;i<=n;i++) val[i]=max(max(i,val[i]),val[i-1]); 
	int now=0;
	dp[now][0]=1;
	for(int i=1;i<=n;i++,now^=1){
		memset(dp[now^1],0,sizeof(dp[0]));
		for(int j=max(0,s[val[i]]-(val[i]-i));j<=s[val[i]];j++)
			dp[now^1][j]=(dp[now][j]+ (j==0?0:dp[now][j-1]) )%mod;
	}
	cout<<dp[now][s[n]]<<endl;
	return 0;
}
```