数论（多个数gcd） CodeForces757B Bash's Big Day

B. Bash's Big Day

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu's Lab. Since Bash is Professor Zulu's favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.

But Zulu warns him that a group of k > 1 Pokemon with strengths {s₁, s₂, s₃, ..., s_k} tend to fight among each other if gcd(s₁, s₂, s₃, ..., s_k) = 1 (see notes for gcd definition).

Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?

Note: A Pokemon cannot fight with itself.

Input

The input consists of two lines.

The first line contains an integer n (1 ≤ n ≤ 10⁵), the number of Pokemon in the lab.

The next line contains n space separated integers, where thei-th of them denotes s_i (1 ≤ s_i ≤ 10⁵), the strength of thei-th Pokemon.

Output

Print single integer — the maximum number of Pokemons Bash can take.

Examples

Input

32 3 4

Output

2

Input

52 3 4 6 7

Output

3

Note

gcd (greatest common divisor) of positive integers set{a₁, a₂, ..., a_n} is the maximum positive integer that divides all the integers{a₁, a₂, ..., a_n}.

In the first sample, we can take Pokemons with strengths{2, 4} since gcd(2, 4) = 2.

In the second sample, we can take Pokemons with strengths{2, 4, 6}, and there is no larger group with gcd ≠ 1.

题意：给一组数，求其中最大公约数为不为1的组合里数字最多的量。

思路：用素数筛法的方法对每个数求因子，son[i]记录·，代码有注释。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int num, son[100000+20];bool cmp(int a, int b) {    return a>b;}int main(){    int n;    while(~scanf("%d", &n)){        memset(son, 0, sizeof(son));        for(int i = 0; i < n; i++) {            scanf("%d", &num);            // 枚举每个数的因子，用son[i]记录i作为因子的次数，最后求出最大的那个            for(int j = 1; j*j <= num; j++) {                if(num % j == 0) {                    son[j]++;                    son[num / j]++;                }                if(j * j == num) {                    son[j]--;                }            }        }//        sort(son+3, son+100000, cmp);//        printf("%d\n", son[2]);        int ans = 1;        for(int i = 2; i < 100020; i++) {            ans = max(ans, son[i]);        }        printf("%d\n", ans);      }    return 0;}