#include <iostream>

using namespace std;

const int MAXN = 3007, MOD = 1000000007;
int R[MAXN], N, M, dp[MAXN][MAXN];
bool S[MAXN];

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> N >> M;
	for (int i = 1; i <= N; ++i) {
		char ch;
		cin >> ch;
		S[i] = ch - '0';
	}
	for (int i = 0; i < M; ++i) {
		int l, r;
		cin >> l >> r;
		R[l] = max(R[l], r);
	}

	dp[0][0] = 1;
	for (int i = 1; i <= N; ++i) {
		R[i] = max(i, max(R[i], R[i - 1]));
		int add = 0;
		for (int j = R[i - 1] + 1; j <= R[i]; ++j) add += S[j];

		for (int j = 0; j <= N; ++j) if (dp[i - 1][j]) {
			if (j + add <= R[i] - i) {
				dp[i][j + add] += dp[i - 1][j];
				if (dp[i][j + add] >= MOD) dp[i][j + add] -= MOD;
			}
			if (j + add - 1 >= 0) {
				dp[i][j + add - 1] += dp[i - 1][j];
				if (dp[i][j + add - 1] >= MOD) dp[i][j + add - 1] -= MOD;
			}
		}
	}

	cout << dp[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 <iostream>

using namespace std;

const int MAXN = 3007, MOD = 1000000007;
int R[MAXN], N, M, dp[MAXN][MAXN];
bool S[MAXN];

int main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> N >> M;
	for (int i = 1; i <= N; ++i) {
		char ch;
		cin >> ch;
		S[i] = ch - '0';
	}
	for (int i = 0; i < M; ++i) {
		int l, r;
		cin >> l >> r;
		R[l] = max(R[l], r);
	}

	dp[0][0] = 1;
	for (int i = 1; i <= N; ++i) {
		R[i] = max(i, max(R[i], R[i - 1]));
		int add = 0;
		for (int j = R[i - 1] + 1; j <= R[i]; ++j) add += S[j];

		for (int j = 0; j <= N; ++j) if (dp[i - 1][j]) {
			if (j + add <= R[i] - i) {
				dp[i][j + add] += dp[i - 1][j];
				if (dp[i][j + add] >= MOD) dp[i][j + add] -= MOD;
			}
			if (j + add - 1 >= 0) {
				dp[i][j + add - 1] += dp[i - 1][j];
				if (dp[i][j + add - 1] >= MOD) dp[i][j + add - 1] -= MOD;
			}
		}
	}

	cout << dp[N][0] << endl;
}
```