#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define pb push_back
#define all(x) begin(x), end(x)
#define SZ(x) (int)(x).size()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int n0 = 2e5 + 123;
const ll t = 1e9;
ll n, a[n0];
int b[n0], c[n0], cnt[20][45];

int main() {
	ios_base::sync_with_stdio(false), cin.tie(NULL);
	cin >> n;
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ld x;
		cin >> x;
		a[i] = round((ld)x * t);
		ll cur = a[i];
		while (cur % 2 == 0) {
			cur /= 2;
			b[i]++;
		}
		cur = a[i];
		while (cur % 5 == 0) {
			cur /= 5;
			c[i]++;
		}
		for (int j = max(0, 18 - c[i]); j < 20; j++) {
			for (int k = max(0, 18 - b[i]); k < 45; k++) {
				ans += cnt[j][k];
			}
		}
		cnt[c[i]][b[i]]++;
	}
	cout << ans;
}

### Prompt

Your challenge is to write a cpp solution to 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;
#define ll long long
#define ld long double
#define pb push_back
#define all(x) begin(x), end(x)
#define SZ(x) (int)(x).size()
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int n0 = 2e5 + 123;
const ll t = 1e9;
ll n, a[n0];
int b[n0], c[n0], cnt[20][45];

int main() {
	ios_base::sync_with_stdio(false), cin.tie(NULL);
	cin >> n;
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ld x;
		cin >> x;
		a[i] = round((ld)x * t);
		ll cur = a[i];
		while (cur % 2 == 0) {
			cur /= 2;
			b[i]++;
		}
		cur = a[i];
		while (cur % 5 == 0) {
			cur /= 5;
			c[i]++;
		}
		for (int j = max(0, 18 - c[i]); j < 20; j++) {
			for (int k = max(0, 18 - b[i]); k < 45; k++) {
				ans += cnt[j][k];
			}
		}
		cnt[c[i]][b[i]]++;
	}
	cout << ans;
}

```