#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int MAX_N = 2e5 + 10;
const ll BASE = 1e9;

int N;
map<pair<int, int>, ll> mp;

int main() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) {
        double _a; scanf("%lf", &_a);
        ll a = _a * BASE + 0.5;
        int cnt2 = 0, cnt5 = 0;
        while (a % 2 == 0) cnt2++, a /= 2;
        while (a % 5 == 0) cnt5++, a /= 5;
        mp[make_pair(cnt2, cnt5)]++;
    }
    ll ans = 0;
    for (auto i : mp) {
        for (auto j : mp) {
            if (i.first <= j.first && i.first.first + j.first.first >= 18 && i.first.second + j.first.second >= 18) {
                if (i.first == j.first) {
                    ans += i.second * (i.second - 1) / 2;
                } else {
                    ans += i.second * j.second;
                }
            }
        }
    }
    printf("%lld\n", ans);
}
### Prompt

Your task is to create 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;

typedef long long ll;

const int MAX_N = 2e5 + 10;
const ll BASE = 1e9;

int N;
map<pair<int, int>, ll> mp;

int main() {
    scanf("%d", &N);
    for (int i = 1; i <= N; i++) {
        double _a; scanf("%lf", &_a);
        ll a = _a * BASE + 0.5;
        int cnt2 = 0, cnt5 = 0;
        while (a % 2 == 0) cnt2++, a /= 2;
        while (a % 5 == 0) cnt5++, a /= 5;
        mp[make_pair(cnt2, cnt5)]++;
    }
    ll ans = 0;
    for (auto i : mp) {
        for (auto j : mp) {
            if (i.first <= j.first && i.first.first + j.first.first >= 18 && i.first.second + j.first.second >= 18) {
                if (i.first == j.first) {
                    ans += i.second * (i.second - 1) / 2;
                } else {
                    ans += i.second * j.second;
                }
            }
        }
    }
    printf("%lld\n", ans);
}
```