#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 3005,  MOD = 1e9+7;
int n, m, l[MX], r[MX], ac[MX], rmax, c;
char s[MX];
ll dp[MX] = {1};
int main(){
	scanf("%d%d\n%s", &n, &m, s);
	rep(i,n) ac[i+1] = ac[i] + s[i] -'0';
	rep(i,m) scanf("%d%d", l+i, r+i);
	rep(i,n){
		for(; c < n && l[c] == i+1; c++) rmax = max(rmax, r[c]);
		if(rmax == i) rmax++;
		for(int j=ac[rmax]-1; j >= max(0,ac[rmax]-rmax+i); j--) (dp[j+1] += dp[j]) %= MOD;
		rep(j,ac[rmax]-rmax+i+1) dp[j] = 0;
	}
	printf("%lld\n", dp[ac[n]]);
}
### Prompt

Create a solution in cpp for 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;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 3005,  MOD = 1e9+7;
int n, m, l[MX], r[MX], ac[MX], rmax, c;
char s[MX];
ll dp[MX] = {1};
int main(){
	scanf("%d%d\n%s", &n, &m, s);
	rep(i,n) ac[i+1] = ac[i] + s[i] -'0';
	rep(i,m) scanf("%d%d", l+i, r+i);
	rep(i,n){
		for(; c < n && l[c] == i+1; c++) rmax = max(rmax, r[c]);
		if(rmax == i) rmax++;
		for(int j=ac[rmax]-1; j >= max(0,ac[rmax]-rmax+i); j--) (dp[j+1] += dp[j]) %= MOD;
		rep(j,ac[rmax]-rmax+i+1) dp[j] = 0;
	}
	printf("%lld\n", dp[ac[n]]);
}
```