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

const int K = 65;
long long cnt[K][K];

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n;
    cin>>n;

    long long ans = 0;

    for (int i=1; i<=n; i++) {
        double d;
        cin>>d;

        long long x = round(d*1000000000);

        int p2 = 0, p5 = 0;
        while (x%2 == 0) x/=2, p2++;
        while (x%5 == 0) x/=5, p5++;
        cnt[p2][p5]++;
        if (p2>=9 && p5>=9) ans--;
    }

    for (int i2=0; i2<K; i2++) {
        for (int i5=0; i5<K; i5++) {
            if (cnt[i2][i5] == 0)   continue;

            for (int j2=0; j2<K; j2++) {
                for (int j5=0; j5<K; j5++) {
                    if (i2+j2>=18 && i5+j5>=18)
                    ans += cnt[i2][i5]*cnt[j2][j5];
                }
            }
        }
    }
    cout<<ans/2<<endl;
}

### 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;

const int K = 65;
long long cnt[K][K];

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);

    int n;
    cin>>n;

    long long ans = 0;

    for (int i=1; i<=n; i++) {
        double d;
        cin>>d;

        long long x = round(d*1000000000);

        int p2 = 0, p5 = 0;
        while (x%2 == 0) x/=2, p2++;
        while (x%5 == 0) x/=5, p5++;
        cnt[p2][p5]++;
        if (p2>=9 && p5>=9) ans--;
    }

    for (int i2=0; i2<K; i2++) {
        for (int i5=0; i5<K; i5++) {
            if (cnt[i2][i5] == 0)   continue;

            for (int j2=0; j2<K; j2++) {
                for (int j5=0; j5<K; j5++) {
                    if (i2+j2>=18 && i5+j5>=18)
                    ans += cnt[i2][i5]*cnt[j2][j5];
                }
            }
        }
    }
    cout<<ans/2<<endl;
}

```