Divisiblity of Differences

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, ..., an (0 ≤ ai ≤ 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 kintegers b1, b2, ..., bk — the selected numbers. If there are multiple possible solutions, print any of them.

Examples

input

Copy

3 2 3
1 8 4

output

Copy

Yes
1 4

input

Copy

3 3 3
1 8 4

output

Copy

No

input

Copy

4 3 5
2 7 7 7

output

Copy

Yes
2 7 7 

题意：给你一个数组，问是否存在k个数，这些数之间任意的差都等于m;
思路：对所有数对m取余，若余数相等，则这些数两两之间差一定能被m整除。

#include <iostream>
#include <bits/stdc++.h>
#define maxn 100005
using namespace std;
    int a[maxn];
int b[maxn];
int vis[maxn];
int main()
{
    int n,k,m,i;
    scanf("%d%d%d",&n,&k,&m);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        b[i]=a[i];
    }
    for(i=1;i<=n;i++)
    {
        b[i]%=m;
    }
    for(i=1;i<=n;i++)
    {
        vis[b[i]]++;
    }
    int flag=0;
    for(i=0;i<m;i++)
    {
        if(vis[i]>=k)
        {
            flag=1;
            printf("Yes
");
            int f=0;
            int cnt=0;
            for(int j=1;j<=n;j++)
            {
                if(b[j]==i)
                {
                    if(f)
                    {
                        printf(" ");
                    }
                    cnt++;
                    printf("%d",a[j]);
                    f=1;
                }
                if(cnt==k) break;
            }
            printf("
");
            break;
        }
    }
    if(flag==0)
    {
        printf("No
");
    }
    return 0;
}