C. Enlarge GCD

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mr. F has nn positive integers, a1,a2,…,ana1,a2,…,an.

He thinks the greatest common divisor of these integers is too small. So he wants to enlarge it by removing some of the integers.

But this problem is too simple for him, so he does not want to do it by himself. If you help him, he will give you some scores in reward.

Your task is to calculate the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.

Input

The first line contains an integer nn (2≤n≤3⋅1052≤n≤3⋅105) — the number of integers Mr. F has.

The second line contains nn integers, a1,a2,…,ana1,a2,…,an (1≤ai≤1.5⋅1071≤ai≤1.5⋅107).

Output

Print an integer — the minimum number of integers you need to remove so that the greatest common divisor of the remaining integers is bigger than that of all integers.

You should not remove all of the integers.

If there is no solution, print «-1» (without quotes).

Examples

input

Copy

3
1 2 4

output

Copy

1

input

Copy

4
6 9 15 30

output

Copy

2

input

Copy

3
1 1 1

output

Copy

-1

Note

In the first example, the greatest common divisor is 11 in the beginning. You can remove 11 so that the greatest common divisor is enlarged to 22. The answer is 11.

In the second example, the greatest common divisor is 33 in the beginning. You can remove 66 and 99 so that the greatest common divisor is enlarged to 1515. There is no solution which removes only one integer. So the answer is 22.

In the third example, there is no solution to enlarge the greatest common divisor. So the answer is −1−1.

題意：當時做的時候不知道用什麼方法做，看了人家的部落格，才知道用 類似於素數篩的做法， 1.5e7 的素篩，也能過，我不會計算複雜度了，哪位看部落格的朋友，給講解一下，在這(❁´ω`❁)謝過了；

先求出 所有數的最大公約數 g，再 g+1 開始篩，請看程式碼：

#include<bits/stdc++.h>
using namespace std;
#define Max (int)(1.5e7+10)
#define ll long long
int book[Max],num[Max];
int gcd(int a,int b)
{
	return (b==0)?a:gcd(b,a%b);
} 
int main()
{
	int n;
	scanf("%d",&n);
	int g = 0,tt;
	for(int i = 0;i<n; i ++ )
	{
		scanf("%d",&tt);
		num[tt] ++;
		if(!i) g = tt;
		else g = gcd(g,tt);
	}
	int ans = n;
	for(int i = g + 1; i<Max;i++)
	{
		if(book[i]) continue;
		int cnt = 0;
		for(int j = i;j < Max; j += i)
		{
			book[j] = 1;
			cnt += num[j];
		}
		ans = min(ans, n - cnt);
	}
	if(ans == n)
		printf("-1\n");
	else printf("%d\n",ans);
	return 0;
}