#include <bits/stdc++.h>
using namespace std;

const int N = 3005;
const int md = 1e9 + 7;

int n, m;
string s;
int r[N], cnt[N], mx[N];
int f[N][N];

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n >> m >> s; s = ' ' + s;
	for (int i = 1; i <= n; ++i) cnt[i] = cnt[i-1] + (s[i] == '1');
	for (int i = 1; i <= m; ++i) {
		int L, R; cin >> L >> R; r[L] = max(r[L], R);
	}
	for (int i = 1; i <= n; ++i) mx[i] = max(mx[i-1], r[i]), mx[i] = max(mx[i], i);

	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) if (f[i-1][j]) {
			int nj = j + max(0, cnt[mx[i]] - cnt[mx[i-1]]);
			int w = mx[i] - i + 1;
			if (w - nj > 0) f[i][nj] = (f[i][nj] + f[i-1][j]) % md;
			if (nj > 0) f[i][nj - 1] = (f[i][nj - 1] + f[i-1][j]) % md;
		}
	}

	cout << f[n][0] << endl;
}
### 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;

const int N = 3005;
const int md = 1e9 + 7;

int n, m;
string s;
int r[N], cnt[N], mx[N];
int f[N][N];

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	cin >> n >> m >> s; s = ' ' + s;
	for (int i = 1; i <= n; ++i) cnt[i] = cnt[i-1] + (s[i] == '1');
	for (int i = 1; i <= m; ++i) {
		int L, R; cin >> L >> R; r[L] = max(r[L], R);
	}
	for (int i = 1; i <= n; ++i) mx[i] = max(mx[i-1], r[i]), mx[i] = max(mx[i], i);

	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) if (f[i-1][j]) {
			int nj = j + max(0, cnt[mx[i]] - cnt[mx[i-1]]);
			int w = mx[i] - i + 1;
			if (w - nj > 0) f[i][nj] = (f[i][nj] + f[i-1][j]) % md;
			if (nj > 0) f[i][nj - 1] = (f[i][nj - 1] + f[i-1][j]) % md;
		}
	}

	cout << f[n][0] << endl;
}
```