Codeforces Round #441 div2 B. 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 

题意 给出n k m接下来有n个数 需从中挑选k个数使得两两之差对m取余等于零

用同余加桶排  如果有一个桶里的数大于等于k就输出Yes和前k个数  没有就输出No

#include <cstdio>#include <cstring>#include <string>#include <iostream>#include <algorithm>#include <cmath>using namespace std;#define ll long long#define mem(a, b) memset(a, b, sizeof(a))#define debug() puts("F*** you everyday!");#define Maxn 100005struct node {        int num, mod;}str[Maxn];int main(){        int n, m, k, x, a[Maxn];        mem(a, 0);        scanf("%d%d%d", &n, &k, &m);        for (int i = 0; i < n; i ++) {                scanf("%d", &str[i].num);                str[i].mod = str[i].num % m;                a[str[i].mod] ++;        }        int u = 0;        for (int i = 0; i < m; i ++) {                if (a[i] >= k) {                        u = 1;                        printf("Yes\n");                        for (int j = 0; j < n && k; j ++) {                                if (str[j].mod == i) {                                        printf("%d ", str[j].num);                                        k --;                                }                        }                        break;                }        }        if (u) {                printf("\n");        }        else {                printf("No\n");        }}