#include <bits/stdc++.h>

const int MAXN = 3010;
const int mod = 1000000007;
void reduce(int & x) { x += x >> 31 & mod; }
char buf[MAXN];
int rs[MAXN], n, m;
int pre[MAXN];
int f[MAXN][MAXN];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m >> buf + 1;
	for (int i = 1; i <= n + 1; ++i) rs[i] = i;
	for (int i = 1, a, b; i <= m; ++i)
		std::cin >> a >> b, rs[a] = std::max(rs[a], b);
	for (int i = 1; i <= n; ++i) rs[i] = std::max(rs[i], rs[i - 1]);
	for (int i = 1; i <= n; ++i)
		pre[i] = pre[i - 1] + (buf[i] & 1);
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) if (int t = f[i - 1][j]) {
			int hav = j + pre[rs[i]] - pre[rs[i - 1]];
			reduce(f[i][hav] += t - mod);
			if (hav) reduce(f[i][hav - 1] += t - mod);
		}
		for (int j = 0; j <= n; ++j)
			if (rs[i] - i - j < 0) f[i][j] = 0;
	}
	std::cout << f[n][0] << std::endl;
	return 0;
}

### 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>

const int MAXN = 3010;
const int mod = 1000000007;
void reduce(int & x) { x += x >> 31 & mod; }
char buf[MAXN];
int rs[MAXN], n, m;
int pre[MAXN];
int f[MAXN][MAXN];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m >> buf + 1;
	for (int i = 1; i <= n + 1; ++i) rs[i] = i;
	for (int i = 1, a, b; i <= m; ++i)
		std::cin >> a >> b, rs[a] = std::max(rs[a], b);
	for (int i = 1; i <= n; ++i) rs[i] = std::max(rs[i], rs[i - 1]);
	for (int i = 1; i <= n; ++i)
		pre[i] = pre[i - 1] + (buf[i] & 1);
	f[0][0] = 1;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) if (int t = f[i - 1][j]) {
			int hav = j + pre[rs[i]] - pre[rs[i - 1]];
			reduce(f[i][hav] += t - mod);
			if (hav) reduce(f[i][hav - 1] += t - mod);
		}
		for (int j = 0; j <= n; ++j)
			if (rs[i] - i - j < 0) f[i][j] = 0;
	}
	std::cout << f[n][0] << std::endl;
	return 0;
}

```