Codeforces Round #441 B. Divisiblity of Differences

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本文介绍了一个算法问题：从一个多集中选择k个整数，使得任意两个数之差都能被m整除。讨论了问题的输入输出格式及解决思路，通过计数并检查余数相同的元素来找出符合条件的k个数。

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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 k integers b1, b2, ..., bk — the selected numbers. If there are multiple possible solutions, print any of them.

Examples

input
3 2 3
1 8 4

output
Yes
1 4 

input
3 3 3
1 8 4

output
No

input
4 3 5
2 7 7 7

output
Yes
2 7 7 

从序列n中挑选k个，使挑选出来的k个被m除后余数相同

打表，然后选取，看是否可以找到

//
//  main.cpp
//  B. Divisiblity of Differences
//
//  Created by wenhan on 2017/10/17.
//  Copyright © 2017年 wenhan. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int a[1000005];
int b[1000005];
int main() {
    int n,k,m;
    scanf("%d%d%d",&n,&k,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    memset(b, 0, sizeof(b));
    for(int i=1;i<=n;i++)
        b[a[i]%m]++;
    int i;
    for(i=0;i<m;i++)
        if(k<=b[i])
            break;
    if(i==m)
        printf("No\n");
    else
    {
        printf("Yes\n");
        int f=0;
        for(int j=1;j<=n;j++)
        {
            if(f==0&&a[j]%m==i)
            {
                f++;
                printf("%d",a[j]);
            }
            else if(a[j]%m==i)
            {
                f++;
                printf(" %d",a[j]);
            }
            if(f==k)
                break;
        }
        printf("\n");
    }
    // insert code here...
    //std::cout << "Hello, World!\n";
    return 0;
}