- Divisiblity of Differences CodeForces

You are given a multiset of n integers. You should select exactly k of them in a such way that the difference between any two of them is divisible by m, or tell that it is impossible.

Numbers can be repeated in the original multiset and in the multiset of selected numbers, but number of occurrences of any number in multiset of selected numbers should not exceed the number of its occurrences in the original multiset.

Input

First line contains three integers n, k and m (2 ≤ k ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) — number of integers in the multiset, number of integers you should select and the required divisor of any pair of selected integers.

Second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁹) — the numbers in the multiset.

Output

If it is not possible to select k numbers in the desired way, output «No» (without the quotes).

Otherwise, in the first line of output print «Yes» (without the quotes). In the second line print k integers b₁, b₂, ..., b_k — the selected numbers. If there are multiple possible solutions, print any of them.

Example

Input

3 2 31 8 4

Output

Yes1 4 

Input

3 3 31 8 4

Output

No

Input

4 3 52 7 7 7

Output

Yes2 7 7 

题意：给你n k m三个整数，再接着为n个整数，问在n中是否含有k个数，他们两两

之间相差为m的倍数，

思路：思路难想，之前用map做，做了半天没写出来，看了学长的代码才想到，我们

可以直接对数据取余，标记出现的次数，代码很简单，一看就会明白。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int num[110000];int book[110000];int main(){    int n,k,m;    while(~scanf("%d %d %d",&n,&k,&m))    {        memset(book,0,sizeof(book));        int x,flag=0;        for(int i=0; i<n; i++)        {            scanf("%d",&num[i]);            int t=num[i]%m;            book[t]++;            if(book[t]>=k)            {                x=t;                flag=1;            }        }        if(flag)        {            int term=0;            printf("Yes\n");            for(int i=0; i<n&&term<k; i++)                if(num[i]%m==x)                {                    if(term==0)                        printf("%d",num[i]);                    else printf(" %d",num[i]);                    term++;                }            printf("\n");        }        else printf("No\n");    }    return 0;}