#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> ii;
typedef long long ll;
ll a[200005];
int tw[200005], fi[200005];
int ct[20][20];
ll ans = 0;
int main(){
    int n;
    scanf("%d",&n);
    for (int i = 0; i <n; i++){
        long double x;
        scanf("%Lf",&x);
        a[i] = (x*1000000000 + 0.5);
        //printf("%lld ",a[i]);
        while (a[i] % 2 == 0){
            tw[i]++;
            a[i] /= 2;
        }
        while (a[i] % 5 == 0){
            fi[i]++;
            a[i] /= 5;
        }
        tw[i] = min(tw[i],18);
        fi[i] = min(fi[i],18);
        for (int k = 18-tw[i]; k <= 18; k++){
            for (int l = 18-fi[i]; l <= 18; l++){
                ans += ct[k][l];
            }
        }
        ct[tw[i]][fi[i]]++;
    }
    printf("%lld\n",ans);
}

### Prompt

In cpp, your task is to solve 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 pair<int,int> ii;
typedef long long ll;
ll a[200005];
int tw[200005], fi[200005];
int ct[20][20];
ll ans = 0;
int main(){
    int n;
    scanf("%d",&n);
    for (int i = 0; i <n; i++){
        long double x;
        scanf("%Lf",&x);
        a[i] = (x*1000000000 + 0.5);
        //printf("%lld ",a[i]);
        while (a[i] % 2 == 0){
            tw[i]++;
            a[i] /= 2;
        }
        while (a[i] % 5 == 0){
            fi[i]++;
            a[i] /= 5;
        }
        tw[i] = min(tw[i],18);
        fi[i] = min(fi[i],18);
        for (int k = 18-tw[i]; k <= 18; k++){
            for (int l = 18-fi[i]; l <= 18; l++){
                ans += ct[k][l];
            }
        }
        ct[tw[i]][fi[i]]++;
    }
    printf("%lld\n",ans);
}

```