#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, mp[25][25], ac[25][25], f, tw, fv;
char c;
long long a, ans;
int main(){
	while(getchar() != '\n');
	while((c = getchar()) != EOF){
		if(c == '\n'){
			for(; a%2==0; a/=2, tw++); if(tw>10)tw=10;
			for(; a%5==0; a/=5, fv++); if(fv>10)fv=10;
			mp[tw+10][fv+10]++;
			a = tw = fv = f = 0;
		}else if(c == '.') f=1;
		else{
			tw -= f; fv -= f;
			a = a*10 + c-'0';
		}
	}
	if(a){
		for(; a%2==0; a/=2, tw++); if(tw>10)tw=10;
		for(; a%5==0; a/=5, fv++); if(fv>10)fv=10;
		mp[tw+10][fv+10]++;
	}
	for(int i=20; i>=0; i--) for(int j=20; j>=0; j--) ac[i][j] = ac[i+1][j] + ac[i][j+1] -ac[i+1][j+1] + mp[i][j];
	rep(i,21) rep(j,21) ans += (long long)mp[i][j] * ac[20-i][20-j];
	printf("%lld\n", (ans-ac[10][10])/2);
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 rep(i,n) for(int i=0; i<(n); i++)
int n, mp[25][25], ac[25][25], f, tw, fv;
char c;
long long a, ans;
int main(){
	while(getchar() != '\n');
	while((c = getchar()) != EOF){
		if(c == '\n'){
			for(; a%2==0; a/=2, tw++); if(tw>10)tw=10;
			for(; a%5==0; a/=5, fv++); if(fv>10)fv=10;
			mp[tw+10][fv+10]++;
			a = tw = fv = f = 0;
		}else if(c == '.') f=1;
		else{
			tw -= f; fv -= f;
			a = a*10 + c-'0';
		}
	}
	if(a){
		for(; a%2==0; a/=2, tw++); if(tw>10)tw=10;
		for(; a%5==0; a/=5, fv++); if(fv>10)fv=10;
		mp[tw+10][fv+10]++;
	}
	for(int i=20; i>=0; i--) for(int j=20; j>=0; j--) ac[i][j] = ac[i+1][j] + ac[i][j+1] -ac[i+1][j+1] + mp[i][j];
	rep(i,21) rep(j,21) ans += (long long)mp[i][j] * ac[20-i][20-j];
	printf("%lld\n", (ans-ac[10][10])/2);
}
```