Codeforces876B-Divisiblity of Differences

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本文探讨了一个数学问题：从一个包含n个整数的多重集中选取k个整数，使得这些整数之间的任意两数之差能被m整除。文章提供了一种解决方案，并附带了实现该方案的伪代码。

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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, ..., 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 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

题意：给你一个长度为n的序列，让你从这n个数里面，选k个数出来，这k个数中任意两个数的差，模m等于零

题解：我们直接将每一个数都mod m。相同的数就是一组。

Code：

var n,k,m,i,num,s:longint;flag:boolean;
    a,q,sum:array[0..200000]of longint;
begin
   readln(n,k,m);
   for i:=1 to n do begin read(a[i]);q[i]:=a[i] mod m;end;
   for i:=1 to n do inc(sum[q[i]]);
   flag:=false;
   for i:=0 to m-1 do if sum[i]>=k then begin flag:=true;break;end;
   if flag then
   begin
      writeln('Yes');
      num:=i;
      for i:=1 to n do
         if q[i]=num then
         begin
            inc(s);
            if s=k then begin writeln(a[i]);break;end
            else write(a[i],' ');
         end;
   end else writeln('No');
end.