CF1047C Enlarge GCD

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

Enlarge GCD

Mr. F has nnn positive integers, a1,a2,⋯ ,ana_1,a_2,\cdots,a_na1,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 n(2≤n≤3⋅105)n (2≤n≤3⋅10^5)n(2≤n≤3⋅105) — the number of integers Mr. F has.

The second line contains nnn integers, a1,a2,⋯ ,an(1≤ai≤1.5⋅107)a_1,a_2,\cdots,a_n (1≤a_i≤1.5⋅10^7)a1,a2,⋯,an(1≤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

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 111 in the beginning. You can remove 111 so that the greatest common divisor is enlarged to 222. The answer is 111.

In the second example, the greatest common divisor is 333 in the beginning. You can remove 666 and 999 so that the greatest common divisor is enlarged to 151515. There is no solution which removes only one integer. So the answer is 222.

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

题解

先把所有数除个gcdgcdgcd，再开个桶记录除gcdgcdgcd后的数，枚举一下所有质数，统计含有该质数的数的个数取maxmaxmax，最后的答案就是n−maxn-maxn−max。

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