Codeforces Round #441 (Div. 2)B. Divisiblity of Differences（哈希的运用）

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 

题意，第一行三个数字N，K,M，接下来N个数，求有没有一个集合大小为k，并且集合内任意两个数mod m的值都相等（一开始以为是每两个数相减后的值mod m相等，然后超时）

因为每两个数mod m都相等，就把他们都存进mod m值的数组中，算法就是哈希。

#include <iostream>#include <stdio.h>#include <string.h>#include <vector>const int maxn=100007;using namespace std;int n,k,m;vector<int> v[maxn];int main(){    cin>>n>>k>>m;    int a;    for(int i=0;i<n;i++)    {        cin>>a;        v[a%m].push_back(a);        if(v[a%m].size()==k)        {            cout<<"Yes"<<endl;            for(int j=0;j<v[a%m].size();j++)            {                cout<<v[a%m][j]<<" ";            }            cout<<endl;            return 0;        }    }    cout<<"No"<<endl;}