#include <bits/stdc++.h>
#define maxn 202333
using namespace std;
long long a[maxn];
int c2[maxn], c5[maxn]; // 50 20
int cnt[60][60];
long long Pow[233] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000};
int main() {
	int n;
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		long long fi, en;
		char tmp[233];
		int ty = scanf("%lld.%s", &fi, tmp);
		if(ty == 1) en = 0, a[i] = fi * 1000000000ll;
		else {
			sscanf(tmp, "%lld", &en);
			int len = strlen(tmp);
			a[i] = fi * 1000000000ll + en * Pow[9 - len];
		}
		while(a[i] % 2 == 0) a[i] /= 2, ++c2[i];
		while(a[i] % 5 == 0) a[i] /= 5, ++c5[i];
	}
	long long ans = 0;
	for(int _ = 1; _ <= n; ++_) {
		int C2 = c2[_], C5 = c5[_];
		if(C2 > 50) C2 = 50;
		if(C5 > 20) C5 = 20;
		for(int i = 18 - min(C2, 18); i <= 50; ++i) {
			for(int j = 18 - min(C5, 18); j <= 20; ++j) {
				ans += cnt[i][j];
			}
		}
		++cnt[C2][C5];
	}
	printf("%lld", ans);
	return 0;
}
### Prompt

In Cpp, your task is to solve 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>
#define maxn 202333
using namespace std;
long long a[maxn];
int c2[maxn], c5[maxn]; // 50 20
int cnt[60][60];
long long Pow[233] = {1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000};
int main() {
	int n;
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		long long fi, en;
		char tmp[233];
		int ty = scanf("%lld.%s", &fi, tmp);
		if(ty == 1) en = 0, a[i] = fi * 1000000000ll;
		else {
			sscanf(tmp, "%lld", &en);
			int len = strlen(tmp);
			a[i] = fi * 1000000000ll + en * Pow[9 - len];
		}
		while(a[i] % 2 == 0) a[i] /= 2, ++c2[i];
		while(a[i] % 5 == 0) a[i] /= 5, ++c5[i];
	}
	long long ans = 0;
	for(int _ = 1; _ <= n; ++_) {
		int C2 = c2[_], C5 = c5[_];
		if(C2 > 50) C2 = 50;
		if(C5 > 20) C5 = 20;
		for(int i = 18 - min(C2, 18); i <= 50; ++i) {
			for(int j = 18 - min(C5, 18); j <= 20; ++j) {
				ans += cnt[i][j];
			}
		}
		++cnt[C2][C5];
	}
	printf("%lld", ans);
	return 0;
}
```