#include <bits/stdc++.h>

int n;
int cnt[60][30];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		double x;
		scanf("%lf", &x);
		long long v = x * 1e9 + 0.5;
		int a = 0, b = 0;
		while (v % 2 == 0) {
			++a;
			v /= 2;
		}
		while (v % 5 == 0) {
			++b;
			v /= 5;
		}
		++cnt[a][b];
	}
	long long ans = 0;
	for (int i = 0; i < 60; ++i) {
		for (int j = 0; j < 30; ++j) {
			for (int k = 0; k < 60; ++k) {
				for (int l = 0; l < 30; ++l) {
					if (i + k >= 18 && j + l >= 18) {
						ans += 1ll * cnt[i][j] * cnt[k][l];
					}
				}
			}
			if (i + i >= 18 && j + j >= 18) {
				ans -= cnt[i][j];
			}
		}
	}
	printf("%lld\n", ans / 2);
}
### Prompt

Create a solution in CPP for the following problem:

You are given N real values A_1, A_2, \ldots, A_N. Compute the number of pairs of indices (i, j) such that i < j and the product A_i \cdot A_j is integer.

Constraints

* 2 \leq N \leq 200\,000
* 0 < A_i < 10^4
* A_i is given with at most 9 digits after the decimal.

Input

Input is given from Standard Input in the following format.


N
A_1
A_2
\vdots
A_N


Output

Print the number of pairs with integer product A_i \cdot A_j (and i < j).

Examples

Input

5
7.5
2.4
17.000000001
17
16.000000000


Output

3


Input

11
0.9
1
1
1.25
2.30000
5
70
0.000000001
9999.999999999
0.999999999
1.000000001


Output

8

### Response

```cpp
#include <bits/stdc++.h>

int n;
int cnt[60][30];

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) {
		double x;
		scanf("%lf", &x);
		long long v = x * 1e9 + 0.5;
		int a = 0, b = 0;
		while (v % 2 == 0) {
			++a;
			v /= 2;
		}
		while (v % 5 == 0) {
			++b;
			v /= 5;
		}
		++cnt[a][b];
	}
	long long ans = 0;
	for (int i = 0; i < 60; ++i) {
		for (int j = 0; j < 30; ++j) {
			for (int k = 0; k < 60; ++k) {
				for (int l = 0; l < 30; ++l) {
					if (i + k >= 18 && j + l >= 18) {
						ans += 1ll * cnt[i][j] * cnt[k][l];
					}
				}
			}
			if (i + i >= 18 && j + j >= 18) {
				ans -= cnt[i][j];
			}
		}
	}
	printf("%lld\n", ans / 2);
}
```