#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxn = 1000 + 10;

int cnt[60][60];

int main(){
	int n;
	cin >> n;
	ll answer = 0;
	for (int i = 0; i < n; i++){
		string s;
		cin >> s;
		ll t = 0;
		int cnT = 9, dec = 0;
		for (int i = 0; i < s.size(); i++){
			if (s[i] == '.'){
				dec = 1;
				continue;
			}
			t = 10LL*t + (int)(s[i]-'0');
			cnT -= dec;
		}
		while (cnT --)
			t = 10LL*t;
		int a = 0, b = 0;
		while (t % 5 == 0){
			t /= 5;
			a ++;
		}
		while (t % 2 == 0){
			t /= 2;
			b ++;
		}
		assert(max(a,b) <= 55);
		for (int i = 0; i <= 55; i++)
			for (int j = 0; j <= 55; j++)
				if (min(i+b,j+a) >= 18)
					answer += cnt[i][j];
		cnt[b][a] ++;
	}
	cout << answer << endl;
}


### 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>
using namespace std;
typedef long long ll;

const int maxn = 1000 + 10;

int cnt[60][60];

int main(){
	int n;
	cin >> n;
	ll answer = 0;
	for (int i = 0; i < n; i++){
		string s;
		cin >> s;
		ll t = 0;
		int cnT = 9, dec = 0;
		for (int i = 0; i < s.size(); i++){
			if (s[i] == '.'){
				dec = 1;
				continue;
			}
			t = 10LL*t + (int)(s[i]-'0');
			cnT -= dec;
		}
		while (cnT --)
			t = 10LL*t;
		int a = 0, b = 0;
		while (t % 5 == 0){
			t /= 5;
			a ++;
		}
		while (t % 2 == 0){
			t /= 2;
			b ++;
		}
		assert(max(a,b) <= 55);
		for (int i = 0; i <= 55; i++)
			for (int j = 0; j <= 55; j++)
				if (min(i+b,j+a) >= 18)
					answer += cnt[i][j];
		cnt[b][a] ++;
	}
	cout << answer << endl;
}


```