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

int main() {
    int n;
    scanf("%d", &n);
    map<pair<int, int>, int> cnt;
    for (int rep = 0; rep < n; ++rep) {
        double x;
        scanf("%lf", &x);
        long long i = llround(x * 1e9);
        int two = 0, five = 0;
        while (i % 2 == 0) {
            ++two;
            i /= 2;
        }

        while (i % 5 == 0) {
            five++;
            i /= 5;
        }
        ++cnt[{two, five}];
    }

    long long 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 += (long long) p1.second * p2.second;
                }else if (p1 == p2) {
                    ans += (long long)p1.second * (p1.second - 1) / 2;
                }
            }
        }
    }
    printf("%lld\n", ans);
}
### Prompt

Please create a solution in cpp 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;

int main() {
    int n;
    scanf("%d", &n);
    map<pair<int, int>, int> cnt;
    for (int rep = 0; rep < n; ++rep) {
        double x;
        scanf("%lf", &x);
        long long i = llround(x * 1e9);
        int two = 0, five = 0;
        while (i % 2 == 0) {
            ++two;
            i /= 2;
        }

        while (i % 5 == 0) {
            five++;
            i /= 5;
        }
        ++cnt[{two, five}];
    }

    long long 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 += (long long) p1.second * p2.second;
                }else if (p1 == p2) {
                    ans += (long long)p1.second * (p1.second - 1) / 2;
                }
            }
        }
    }
    printf("%lld\n", ans);
}
```