#include <cstdio>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
using namespace std;

const int mod = 1000000007;
const int Maxn = 3005;

int n, m;
string S;
int ones[Maxn];
int mx[Maxn];
int dp[Maxn][Maxn];

int main()
{
	scanf("%d %d", &n, &m);
	cin >> S;
	for (int i = 0; i < n; i++) {
		ones[i] = (i? ones[i - 1]: 0) + (S[i] == '1');
		mx[i] = i;
	}
	while (m--) {
		int l, r; scanf("%d %d", &l, &r); l--; r--;
		for (int i = l; i <= r; i++)
			mx[i] = max(mx[i], r);
	}
	dp[0][0] = 1;
	for (int i = 0; i < n; i++)
		for (int j = 0; j <= i; j++) if (dp[i][j] > 0) {
			int lones = ones[mx[i]] - j;
			int lzeros = mx[i] - i + 1 - lones;
			if (lones > 0) dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j]) % mod;
			if (lzeros > 0) dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % mod;
		}
	printf("%d\n", dp[n][ones[n - 1]]);
	return 0;
}
### Prompt

Please formulate 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 <cstdio>
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
using namespace std;

const int mod = 1000000007;
const int Maxn = 3005;

int n, m;
string S;
int ones[Maxn];
int mx[Maxn];
int dp[Maxn][Maxn];

int main()
{
	scanf("%d %d", &n, &m);
	cin >> S;
	for (int i = 0; i < n; i++) {
		ones[i] = (i? ones[i - 1]: 0) + (S[i] == '1');
		mx[i] = i;
	}
	while (m--) {
		int l, r; scanf("%d %d", &l, &r); l--; r--;
		for (int i = l; i <= r; i++)
			mx[i] = max(mx[i], r);
	}
	dp[0][0] = 1;
	for (int i = 0; i < n; i++)
		for (int j = 0; j <= i; j++) if (dp[i][j] > 0) {
			int lones = ones[mx[i]] - j;
			int lzeros = mx[i] - i + 1 - lones;
			if (lones > 0) dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j]) % mod;
			if (lzeros > 0) dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % mod;
		}
	printf("%d\n", dp[n][ones[n - 1]]);
	return 0;
}
```