#include <bits/stdc++.h>
#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;

const int M = 64;

int main() {
    int N; cin >> N;
    vector<vector<int64_t>> A(M, vector<int64_t>(M));

    rep(i, N) {
        long double a; cin >> a;
        int64_t n = (int64_t)(a * 1e9 + 0.5);
        int two = 0, five = 0;
        while(n % 2 == 0) n /= 2, two++;
        while(n % 5 == 0) n /= 5, five++;
        A[two][five]++;
    }

    vector<pair<int, int>> pairs;
    rep(i, M) rep(j, M) pairs.emplace_back(i, j);

    int64_t ans = 0;
    rep(i, pairs.size()) for(int j = i; j < pairs.size(); ++j) {
        auto[t, f] = pairs[i];
        auto[_t, _f] = pairs[j];
        if(t + _t < 18 || f + _f < 18) continue;
        if(i == j) ans += A[t][f] * (A[t][f] - 1) / 2;
        else ans += A[t][f] * A[_t][_f];
    }
    cout << ans << '\n';
    return 0;
}

### 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>
#define rep(i, n) for(int i = 0; i < (n); ++i)
using namespace std;

const int M = 64;

int main() {
    int N; cin >> N;
    vector<vector<int64_t>> A(M, vector<int64_t>(M));

    rep(i, N) {
        long double a; cin >> a;
        int64_t n = (int64_t)(a * 1e9 + 0.5);
        int two = 0, five = 0;
        while(n % 2 == 0) n /= 2, two++;
        while(n % 5 == 0) n /= 5, five++;
        A[two][five]++;
    }

    vector<pair<int, int>> pairs;
    rep(i, M) rep(j, M) pairs.emplace_back(i, j);

    int64_t ans = 0;
    rep(i, pairs.size()) for(int j = i; j < pairs.size(); ++j) {
        auto[t, f] = pairs[i];
        auto[_t, _f] = pairs[j];
        if(t + _t < 18 || f + _f < 18) continue;
        if(i == j) ans += A[t][f] * (A[t][f] - 1) / 2;
        else ans += A[t][f] * A[_t][_f];
    }
    cout << ans << '\n';
    return 0;
}

```