Codeforces876B-Divisiblity of Differences

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 k integers b1, b2, ..., bk —
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.