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
题意：给出 n个数，问，是否可以选出k个数，使得这k个数两两之差为m的倍数。
思路：如个x-y是m的倍数，那么，x，y只有满足 x=a+o×m，y=a+p×m。（o，p属于正整数）x-y才可能是m的倍数。
那么，只要对每个数取余，看余数相等的数是否可以满足k即可。
#include "iostream"
#include "map"
using namespace std;
map<int,int> mp;
const int Max=1e5+10;
int a[Max];
int main()
{
ios::sync_with_stdio(false);
int n,k,m,x,flag=0;
mp.clear();
cin>>n>>k>>m;
for(int i=0;i<n;i++){
cin>>a[i];
mp[a[i]%m]++;
}
for(int i=0;i<m;i++)
if(mp[i]>=k&&!flag){
flag=1;
cout<<"Yes"<<endl;
for(int j=0;j<n;j++)
if(a[j]%m==i&&k) {cout<<a[j]<<" ";k--;}
}
if(!flag) cout<<"No";
cout<<endl;
return 0;
}