#include<bits/stdc++.h>

using namespace std;

typedef long long llint;
typedef pair <int, int> pi;

llint n, sol;
llint br[60][60];

int f (llint val, int p) {
    int cnt = 0;
    while (val % p == 0) {
        val /= p;
        cnt++;
    }
    return cnt;
}

int main () {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) {
        long double x;
        cin >> x;
        x *= 1e9;
        llint val = round(x);
        int a = f(val, 2);
        int b = f(val, 5);
        br[a][b]++;
        if (a >= 9 && b >= 9) sol--;
    }
    for (int i = 0; i < 60; i++) {
        for (int j = 0; j < 60; j++) {
            for (int k = 0; k < 60; k++) {
                for (int h = 0; h < 60; h++) {
                    if (i + k >= 18 && j + h >= 18) sol += br[i][j] * br[k][h];
                }
            }
        }
    }
    cout << sol / 2;
    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>

using namespace std;

typedef long long llint;
typedef pair <int, int> pi;

llint n, sol;
llint br[60][60];

int f (llint val, int p) {
    int cnt = 0;
    while (val % p == 0) {
        val /= p;
        cnt++;
    }
    return cnt;
}

int main () {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) {
        long double x;
        cin >> x;
        x *= 1e9;
        llint val = round(x);
        int a = f(val, 2);
        int b = f(val, 5);
        br[a][b]++;
        if (a >= 9 && b >= 9) sol--;
    }
    for (int i = 0; i < 60; i++) {
        for (int j = 0; j < 60; j++) {
            for (int k = 0; k < 60; k++) {
                for (int h = 0; h < 60; h++) {
                    if (i + k >= 18 && j + h >= 18) sol += br[i][j] * br[k][h];
                }
            }
        }
    }
    cout << sol / 2;
    return 0;
}

```