#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;

typedef long long ll;
int n, m;

double A[maxn];
ll B[maxn];

int main() {

    cin >> n;
    map<pair<int, int>, int> cnt;
    for (int i = 1;i <= n; ++i) {
        cin >> A[i];
        A[i] *= 1e9;
        B[i] = llround(A[i]);

        int two = 0, five = 0;
        while (B[i] % 2 == 0) two++, B[i] /= 2; 
        while (B[i] % 5 == 0) five++, B[i] /= 5; 
        cnt[{two, five}] ++;
    }
    ll ans = 0;
    for (auto p1 : cnt) {
        for (auto p2 : cnt) {
            if (p1.first.first-9 + p2.first.first-9 >= 0 && p1.first.second-9 + p2.first.second-9 >= 0)
                if (p1 < p2) ans += (ll) p1.second * p2.second;
                else if (p1 == p2) ans += (ll) p1.second * (p1.second - 1)/2;
        }
    }
    cout << ans << endl;
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 maxn = 2e5 + 5;

typedef long long ll;
int n, m;

double A[maxn];
ll B[maxn];

int main() {

    cin >> n;
    map<pair<int, int>, int> cnt;
    for (int i = 1;i <= n; ++i) {
        cin >> A[i];
        A[i] *= 1e9;
        B[i] = llround(A[i]);

        int two = 0, five = 0;
        while (B[i] % 2 == 0) two++, B[i] /= 2; 
        while (B[i] % 5 == 0) five++, B[i] /= 5; 
        cnt[{two, five}] ++;
    }
    ll ans = 0;
    for (auto p1 : cnt) {
        for (auto p2 : cnt) {
            if (p1.first.first-9 + p2.first.first-9 >= 0 && p1.first.second-9 + p2.first.second-9 >= 0)
                if (p1 < p2) ans += (ll) p1.second * p2.second;
                else if (p1 == p2) ans += (ll) p1.second * (p1.second - 1)/2;
        }
    }
    cout << ans << endl;
    return 0;
}
```