#include <cstdio>
char s[64];
int a[64][64];
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		int t;
		for (int j = 0; j < 9; j++) {
			s[j] = '0';
		}
		scanf("%d.%s", &t, s);
		long long k = t * 1000000000ll;
		t = 0;
		for (int j = 0; j < 9; j++) {
			if (!s[j]) {
				t = t * 10;
			} else {
				t = t * 10 + s[j] - '0';
			}
		}
		k += t;
		int c2 = 0, c5 = 0;
		while (k % 2 == 0) {
			k /= 2;
			c2++;
		}
		while (k % 5 == 0) {
			k /= 5;
			c5++;
		}
		a[c5][c2]++;
	}
	long long ans = 0;
	for (int i = 0; i < 64; i++) {
		for (int j = 0; j < 64; j++) {
			for (int k = 0; k < 64; k++) {
				for (int l = 0; l < 64; l++) {
					if (i + k >= 18 && j + l >= 18) ans += (long long)a[i][j] * a[k][l];
				}
			}
			if (i + i >= 18 && j + j >= 18) ans -= a[i][j];
		}
	}
	printf("%lld\n", ans / 2);
	return 0;
}

### Prompt

Develop a solution in cpp 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 <cstdio>
char s[64];
int a[64][64];
int main() {
	int n;
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		int t;
		for (int j = 0; j < 9; j++) {
			s[j] = '0';
		}
		scanf("%d.%s", &t, s);
		long long k = t * 1000000000ll;
		t = 0;
		for (int j = 0; j < 9; j++) {
			if (!s[j]) {
				t = t * 10;
			} else {
				t = t * 10 + s[j] - '0';
			}
		}
		k += t;
		int c2 = 0, c5 = 0;
		while (k % 2 == 0) {
			k /= 2;
			c2++;
		}
		while (k % 5 == 0) {
			k /= 5;
			c5++;
		}
		a[c5][c2]++;
	}
	long long ans = 0;
	for (int i = 0; i < 64; i++) {
		for (int j = 0; j < 64; j++) {
			for (int k = 0; k < 64; k++) {
				for (int l = 0; l < 64; l++) {
					if (i + k >= 18 && j + l >= 18) ans += (long long)a[i][j] * a[k][l];
				}
			}
			if (i + i >= 18 && j + j >= 18) ans -= a[i][j];
		}
	}
	printf("%lld\n", ans / 2);
	return 0;
}

```