#include<cstdio>
const int sz = 50;
int n;
char s[110];
int d[200100][2];
int cnt[110][110];
void scanreal(int idx) {
	scanf("%s", s);
	long long int val = 0;
	int flag = 0;
	int divis = 0;
	for (int i = 0; s[i]; i++) {
		if (s[i] == '.') {
			flag = 1;
		}
		else {
			val *= 10;
			val += s[i] - '0';
			if (flag == 1)divis++;
		}
	}
	d[idx][0] = -divis;
	while (val % 2 == 0) {
		d[idx][0]++;
		val /= 2;
	}
	d[idx][1] = -divis;
	while (val % 5 == 0) {
		d[idx][1]++;
		val /= 5;
	}
	cnt[d[idx][0] + sz][d[idx][1] + sz]++;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanreal(i);
	}
	long long int ans = 0;
	for (int i = 0; i < 2 * sz; i++) {
		for (int j = 0; j < 2 * sz; j++) {
			if (i >= sz && j >= sz)ans -= cnt[i][j];
			for (int ii = 2 * sz - i; ii < 2 * sz; ii++) {
				for (int jj = 2 * sz - j; jj < 2 * sz; jj++) {
					ans += 1LL * cnt[i][j] * cnt[ii][jj];
				}
			}
		}
	}
	ans /= 2;
	printf("%lld\n", ans);
	return 0;
}

### Prompt

Please create a solution in Cpp 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<cstdio>
const int sz = 50;
int n;
char s[110];
int d[200100][2];
int cnt[110][110];
void scanreal(int idx) {
	scanf("%s", s);
	long long int val = 0;
	int flag = 0;
	int divis = 0;
	for (int i = 0; s[i]; i++) {
		if (s[i] == '.') {
			flag = 1;
		}
		else {
			val *= 10;
			val += s[i] - '0';
			if (flag == 1)divis++;
		}
	}
	d[idx][0] = -divis;
	while (val % 2 == 0) {
		d[idx][0]++;
		val /= 2;
	}
	d[idx][1] = -divis;
	while (val % 5 == 0) {
		d[idx][1]++;
		val /= 5;
	}
	cnt[d[idx][0] + sz][d[idx][1] + sz]++;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanreal(i);
	}
	long long int ans = 0;
	for (int i = 0; i < 2 * sz; i++) {
		for (int j = 0; j < 2 * sz; j++) {
			if (i >= sz && j >= sz)ans -= cnt[i][j];
			for (int ii = 2 * sz - i; ii < 2 * sz; ii++) {
				for (int jj = 2 * sz - j; jj < 2 * sz; jj++) {
					ans += 1LL * cnt[i][j] * cnt[ii][jj];
				}
			}
		}
	}
	ans /= 2;
	printf("%lld\n", ans);
	return 0;
}

```