#include <iostream>
#include <string>
#include <vector>
using namespace std;

typedef vector<int> vi;

const int MOD = (int)1e9 + 7;
int N, M;
string S;
vi sum1, lr;
vector<vi> dp;

int main(void) {
	int l, r;

	cin >> N >> M >> S;

	sum1.resize(N + 1, 0);
	lr.resize(N + 1, 0);

	for (int i = 1; i <= N; ++i) {
		sum1[i] = (S[i - 1] == '1') ? sum1[i - 1] + 1 : sum1[i - 1];
		lr[i] = i;
	}

	for (int i = 0; i < M; ++i) {
		cin >> l >> r;
		for (int j = l; j <= r; ++j) {
			if (r > lr[j])
				lr[j] = r;
		}
	}

	vi v(N + 1, 0);
	dp.resize(N + 1, v);
	dp[0][0] = 1;
	for (int i = 0; i < N; ++i) {
		l = i - (lr[i] - sum1[lr[i]]);
		if (l < 0)
			l = 0;
		r = sum1[lr[i]];
		for (int j = l; j <= r; ++j) {
			dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
			dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j]) % MOD;
		}
	}

	int ans = dp[N][sum1[N]];
	cout << ans << '\n';
	return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
#include <string>
#include <vector>
using namespace std;

typedef vector<int> vi;

const int MOD = (int)1e9 + 7;
int N, M;
string S;
vi sum1, lr;
vector<vi> dp;

int main(void) {
	int l, r;

	cin >> N >> M >> S;

	sum1.resize(N + 1, 0);
	lr.resize(N + 1, 0);

	for (int i = 1; i <= N; ++i) {
		sum1[i] = (S[i - 1] == '1') ? sum1[i - 1] + 1 : sum1[i - 1];
		lr[i] = i;
	}

	for (int i = 0; i < M; ++i) {
		cin >> l >> r;
		for (int j = l; j <= r; ++j) {
			if (r > lr[j])
				lr[j] = r;
		}
	}

	vi v(N + 1, 0);
	dp.resize(N + 1, v);
	dp[0][0] = 1;
	for (int i = 0; i < N; ++i) {
		l = i - (lr[i] - sum1[lr[i]]);
		if (l < 0)
			l = 0;
		r = sum1[lr[i]];
		for (int j = l; j <= r; ++j) {
			dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % MOD;
			dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j]) % MOD;
		}
	}

	int ans = dp[N][sum1[N]];
	cout << ans << '\n';
	return 0;
}

```