#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;

const ll oo = 1e18;
const int MOD = 1000000007;
const int N = 500002;
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define cpx complex<double>

int n , m , mx[3010], dp[3010][3010], pre[3010];
string s;

int calc(int l , int r , int ones){
	if(ones < 0)return 0;
	if(l == n)return (ones == 0?1:0);
	if(l > r && ones != 0)return 0;
	if(r >= l && ones > (r-l+1))return 0;
	if(mx[l]>r){
		ones += pre[mx[l]]-(r==-1?0:pre[r]);
		r = mx[l];
	}
	int &ret = dp[l][ones];
	if(ret != -1)return ret;
	return ret = (1ll*calc(l+1,r,ones)+calc(l+1,r,ones-1))%MOD;
}
int main() {
	memset(dp,-1,sizeof dp);
	cin >> n >> m >> s;
	for (int i = 0 ; i < n ; ++i) {
		pre[i] = s[i] - '0';
		mx[i] = i;
		if (i)pre[i] += pre[i - 1];
	}
	for (int l , r , i = 0 ; i < m ; ++i) {
		cin >> l >> r;
		l--,r--;
		mx[l] = max(mx[l], r);
	}
	printf("%d\n", calc(0,-1,0));
	return 0;
}
### Prompt

Generate 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;
typedef pair<int, int> pii;

const ll oo = 1e18;
const int MOD = 1000000007;
const int N = 500002;
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define cpx complex<double>

int n , m , mx[3010], dp[3010][3010], pre[3010];
string s;

int calc(int l , int r , int ones){
	if(ones < 0)return 0;
	if(l == n)return (ones == 0?1:0);
	if(l > r && ones != 0)return 0;
	if(r >= l && ones > (r-l+1))return 0;
	if(mx[l]>r){
		ones += pre[mx[l]]-(r==-1?0:pre[r]);
		r = mx[l];
	}
	int &ret = dp[l][ones];
	if(ret != -1)return ret;
	return ret = (1ll*calc(l+1,r,ones)+calc(l+1,r,ones-1))%MOD;
}
int main() {
	memset(dp,-1,sizeof dp);
	cin >> n >> m >> s;
	for (int i = 0 ; i < n ; ++i) {
		pre[i] = s[i] - '0';
		mx[i] = i;
		if (i)pre[i] += pre[i - 1];
	}
	for (int l , r , i = 0 ; i < m ; ++i) {
		cin >> l >> r;
		l--,r--;
		mx[l] = max(mx[l], r);
	}
	printf("%d\n", calc(0,-1,0));
	return 0;
}
```