#include "bits/stdc++.h"

//#include <atcoder/all>

using namespace std;
//using namespace atcoder;

//#define int long long
#define ll long long

ll INF = (1LL << 60);
int mod = 998244353;
using P = pair<int, int>;



signed main() {
	int N; cin >> N;
	vector<ll>A(N);
	for (int i = 0; i < N; i++) {
		double a; cin >> a;
		A[i] = (ll)llround(a * 1e9);
	}
	map<P, int>cnt;
	for (int i = 0; i < N; i++) {
		int cnt2 = 0, cnt5 = 0;
		while (A[i] % 2 == 0) {
			cnt2++;
			A[i] /= 2;
		}
		while (A[i] % 5 == 0) {
			cnt5++;
			A[i] /= 5;
		}
		cnt[{cnt2, cnt5}]++;
	}
	ll ans = 0;
	for (auto p1 : cnt) {
		for (auto p2 : cnt) {
			if (p1.first.first + p2.first.first >= 18 && p1.first.second + p2.first.second >= 18) {
				if (p1 < p2) {
					ans += (ll)p1.second*p2.second;
				}
				else if(p1 == p2){
					ans += (ll)p1.second*(p1.second - 1) / 2;
				}
			}

		}
	}
	cout << ans << endl;
	return 0;
}

### 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"

//#include <atcoder/all>

using namespace std;
//using namespace atcoder;

//#define int long long
#define ll long long

ll INF = (1LL << 60);
int mod = 998244353;
using P = pair<int, int>;



signed main() {
	int N; cin >> N;
	vector<ll>A(N);
	for (int i = 0; i < N; i++) {
		double a; cin >> a;
		A[i] = (ll)llround(a * 1e9);
	}
	map<P, int>cnt;
	for (int i = 0; i < N; i++) {
		int cnt2 = 0, cnt5 = 0;
		while (A[i] % 2 == 0) {
			cnt2++;
			A[i] /= 2;
		}
		while (A[i] % 5 == 0) {
			cnt5++;
			A[i] /= 5;
		}
		cnt[{cnt2, cnt5}]++;
	}
	ll ans = 0;
	for (auto p1 : cnt) {
		for (auto p2 : cnt) {
			if (p1.first.first + p2.first.first >= 18 && p1.first.second + p2.first.second >= 18) {
				if (p1 < p2) {
					ans += (ll)p1.second*p2.second;
				}
				else if(p1 == p2){
					ans += (ll)p1.second*(p1.second - 1) / 2;
				}
			}

		}
	}
	cout << ans << endl;
	return 0;
}

```