Your task is to find such array tt that you can cut out the copy of tt from ss maximum number of times. If there are multiple answers, you may choose any of them.

The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of elements in ss and the desired number of elements in tt, respectively.

The second line of the input contains exactly nn integers s1,s2,…,sns1,s2,…,sn (1≤si≤2⋅1051≤si≤2⋅105).

Print kk integers — the elements of array tt such that you can cut out maximum possible number of copies of this array from ss. If there are multiple answers, print any of them. The required array tt can contain duplicate elements. All the elements of tt (t1,t2,…,tkt1,t2,…,tk) should satisfy the following condition: 1≤ti≤2⋅1051≤ti≤2⋅105.

7 3
1 2 3 2 4 3 1

1 2 3 

10 4
1 3 1 3 10 3 7 7 12 3

7 3 1 3

15 2
1 2 1 1 1 2 1 1 2 1 2 1 1 1 1

1 1 

The first example is described in the problem statement.

In the second example the only answer is [7,3,1,3][7,3,1,3] and any its permutations. It can be shown that you cannot choose any other array such that the maximum number of copies you can cut out would be equal to 22.

In the third example the array tt can be cut out 55 times.

程式碼：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 2e5 + 10;
int n, k;
int a[maxn];
int times[maxn];

bool can(int x) {
    int cnt = 0;
    for(int i = 1; i < maxn; i ++)
        cnt += times[i] / x;
    if(cnt >= k) return true;
    return false;
}

int main() {
    memset(times, 0, sizeof(times));
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; i ++) {
        scanf("%d", &a[i]);
        times[a[i]] ++;
    }

    int l = 1, r = n, mid, maxx = 1;
    while(l <= r) {
        mid = (r - l) / 2 + l;
        if(can(mid)) maxx = mid, l = mid + 1;
        else r = mid - 1;
    }

    int ans[maxn];
    int rec = 0;
    for(int i = 1; i < maxn; i ++) {
        if(times[i] >= maxx)
        for(int j = 1; j <= times[i] / maxx; j ++) {
            ans[rec ++] = i;
            if(rec == k) break;
        }
    }

    for(int i = 0; i < k; i ++) {
        printf("%d", ans[i]);
        printf("%s", i != k - 1 ? " " : "\n");
    }
    return 0;
}

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