CF1047C Enlarge GCD

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本文解析了CodeForces竞赛中一道名为“EnlargeGCD”的C级题目，介绍了如何通过去除部分整数来增大剩余整数的最大公约数的算法思路。文章详细解释了算法步骤，包括预处理、质数筛法以及复杂度分析。

原题链接：http://codeforces.com/contest/1047/problem/C

Enlarge GCD

Mr. F has $n$ positive integers, $a_1,a_2,\cdots,a_n$ .

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 $n (2≤n≤3⋅10^5)$ — the number of integers Mr. F has.

The second line contains $n$ integers, $a_1,a_2,\cdots,a_n (1≤a_i≤1.5⋅10^7)$ .

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

3
1 2 4

output

input

4
6 9 15 30

output

input

3
1 1 1

output

-1

Note

In the first example, the greatest common divisor is $1$ in the beginning. You can remove $1$ so that the greatest common divisor is enlarged to $2$ . The answer is $1$ .

In the second example, the greatest common divisor is $3$ in the beginning. You can remove $6$ and $9$ so that the greatest common divisor is enlarged to $15$ . There is no solution which removes only one integer. So the answer is $2$ .

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

题解

先把所有数除个 $g c d$ ，再开个桶记录除 $g c d$ 后的数，枚举一下所有质数，统计含有该质数的数的个数取 $m a x$ ，最后的答案就是 $n - m a x$ 。

因为质数的个数大约为 $\frac{n}{\log n}$ 个，枚举倍数的复杂度大约也是 $\log n$ 的，所以总复杂度 $O (n)$ 。

代码

#include<bits/stdc++.h>
using namespace std;
const int M=2e7+5,N=3e5+5;
int cot[M],val[N],n,g,mx,ans;
bool vis[M];
void in(){scanf("%d",&n);}
void ac()
{
	for(int i=1;i<=n;++i)scanf("%d",&val[i]),g=__gcd(g,val[i]);
	for(int i=1;i<=n;++i)++cot[mx=max(val[i]/g,mx),val[i]/g];
	for(int i=2,j,s;i<=mx;++i,ans=max(ans,s))
	if(!vis[i])for(j=i,s=0;j<=mx;j+=i)s+=cot[j],vis[j]=1;
	printf("%d",ans?n-ans:-1);
}
int main(){in();ac();}