Codeforces 876B. Divisiblity of Differences

B. Divisiblity of Differences

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

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 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 

题目大意：给你n个数，在这n个数中选出k个数，使他们满足任意的差都能整除m

这道题挺水，可是我没能写出来......后来想想真的很水.....

我们可以用vector来储存余数相同的几个数字，输入的时候每个数字都先对m取模，只要是同余数的，相减之后的差肯定是能整除m的

#include<iostream>#include<vector>using namespace std;const int maxn=1e5+7;vector<int> vec[maxn];int main(){int n,k,m;cin>>n>>k>>m;for(int i=0;i<n;i++){int a;cin>>a;vec[a%m].push_back(a);}bool flag=true;for(int i=0;i<m;i++){if(vec[i].size()>=k){cout<<"Yes"<<endl;for(int j=0;j<k;j++){cout<<vec[i][j]<<' ';}cout<<endl;flag=false;break;}}if(flag) cout<<"No"<<endl;}