#include <bits/stdc++.h>
using namespace std;
#define M 3010
#define MOD 1000000007
char s[M]; int sum[M];
int r[M];
int f[M][M];
int main() {
	int n, m; scanf("%d %d", &n, &m);
	scanf("%s", s + 1);
	for (int i = 1; i <= n; i ++) sum[i] = sum[i - 1] + (s[i] == '1');
	for (int i = 1; i <= n; i ++) r[i] = i;
	for (int i = 1; i <= m; i ++) {
		int x, y; scanf("%d %d", &x, &y);
		r[x] = max(r[x], y);
	}
	for (int i = 1; i <= n; i ++) r[i] = max(r[i - 1], r[i]);
	f[0][0] = 1;
	for (int i = 1; i <= n; i ++) {
		for (int j = 0; j <= sum[r[i - 1]]; j ++)
			if(f[i - 1][j]) {
				if(sum[r[i]] > j) (f[i][j + 1] += f[i - 1][j]) %= MOD;
				if(sum[r[i]] - j < r[i] - i + 1) (f[i][j] += f[i - 1][j]) %= MOD;
			}
	}
	cout << f[n][sum[n]];
	return 0;
}
### Prompt

Your task is to create 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;
#define M 3010
#define MOD 1000000007
char s[M]; int sum[M];
int r[M];
int f[M][M];
int main() {
	int n, m; scanf("%d %d", &n, &m);
	scanf("%s", s + 1);
	for (int i = 1; i <= n; i ++) sum[i] = sum[i - 1] + (s[i] == '1');
	for (int i = 1; i <= n; i ++) r[i] = i;
	for (int i = 1; i <= m; i ++) {
		int x, y; scanf("%d %d", &x, &y);
		r[x] = max(r[x], y);
	}
	for (int i = 1; i <= n; i ++) r[i] = max(r[i - 1], r[i]);
	f[0][0] = 1;
	for (int i = 1; i <= n; i ++) {
		for (int j = 0; j <= sum[r[i - 1]]; j ++)
			if(f[i - 1][j]) {
				if(sum[r[i]] > j) (f[i][j + 1] += f[i - 1][j]) %= MOD;
				if(sum[r[i]] - j < r[i] - i + 1) (f[i][j] += f[i - 1][j]) %= MOD;
			}
	}
	cout << f[n][sum[n]];
	return 0;
}
```