#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double

int main() {
    int N;
    cin >> N;
    map<pair<int, int>, ll> mp;
    ll ans = 0;
    for (int i = 0; i < N; i++) {
        string S;
        cin >> S;
        ll t = 0;
        ll x = 0;
        for (int j = 0; j < (int)S.size(); j++) {
            if (S[j] == '.') {
                t = (int)S.size()-j-1;
            }else {
                x = x * 10 + (int)(S[j]-'0');
            }
        }
        ll pow2 = -t, pow5 = -t;
        while (x%2 == 0){
            x /= 2;
            pow2++;
        }
        while (x%5 == 0){
            x /= 5;
            pow5++;
        }
        for (auto p : mp) {
            int m2 = p.first.first, m5 = p.first.second;
            if (m2+pow2 >= 0 && pow5+m5 >= 0) ans += p.second;
        }
        mp[{pow2, pow5}]++;
    }
    cout << ans << endl;
    return 0;
}

### 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;
#define ll long long
#define ld long double

int main() {
    int N;
    cin >> N;
    map<pair<int, int>, ll> mp;
    ll ans = 0;
    for (int i = 0; i < N; i++) {
        string S;
        cin >> S;
        ll t = 0;
        ll x = 0;
        for (int j = 0; j < (int)S.size(); j++) {
            if (S[j] == '.') {
                t = (int)S.size()-j-1;
            }else {
                x = x * 10 + (int)(S[j]-'0');
            }
        }
        ll pow2 = -t, pow5 = -t;
        while (x%2 == 0){
            x /= 2;
            pow2++;
        }
        while (x%5 == 0){
            x /= 5;
            pow5++;
        }
        for (auto p : mp) {
            int m2 = p.first.first, m5 = p.first.second;
            if (m2+pow2 >= 0 && pow5+m5 >= 0) ans += p.second;
        }
        mp[{pow2, pow5}]++;
    }
    cout << ans << endl;
    return 0;
}

```