Diverse Team

A. Diverse Team

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are $n$

students in a school class, the rating of the $i$-th student on Codehorses is $a_i$. You have to form a team consisting of $k$ students ($1\le k\le n$

) such that the ratings of all team members are distinct.

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$

distinct numbers which should be the indices of students in the team you form. If there are multiple answers, print any of them.

Input

The first line contains two integers $n$

and $k$ ($1\le k\le n\le 100$

) — the number of students and the size of the team you have to form.

The second line contains $n$

integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 100$), where $a_i$ is the rating of $i$

-th student.

Output

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print $k$

distinct integers from $1$ to $n$

which should be the indices of students in the team you form. All the ratings of the students in the team should be distinct. You may print the indices in any order. If there are multiple answers, print any of them.

Assume that the students are numbered from $1$

to $n$

.

Examples

Input

Copy

5 3
15 13 15 15 12

Output

Copy

YES
1 2 5 

Input

Copy

5 4
15 13 15 15 12

Output

Copy

NO

Input

Copy

4 4
20 10 40 30

Output

Copy

YES
1 2 3 4 

Note

All possible answers for the first example:

• {1 2 5}
• {2 3 5}
• {2 4 5}

Note that the order does not matter.

 

#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>

#define mod 1000000007
const int INF = 0x3f3f3f3f;
typedef long long ll;
const int maxn = 15;
using namespace std;

int n, k;
bool flag;
int ans[105];
bool vis[105];

int main()
{
    memset(vis, 0, sizeof(vis));
    scanf("%d%d", &n, &k);
    int cnt = 0;
    for(int i = 0; i < n; i++)
    {
        int temp;
        scanf("%d", &temp);
        if(!vis[temp])
        {
            vis[temp] = true;
            ans[cnt++] = i;
        }
    }
    if(cnt >= k)
    {
        printf("YES\n");
        for(int i = 0; i < k ; i++)
        {
            if(i == k-1)
            {
                printf("%d\n", ans[i]+1);
                continue;
            }
            printf("%d ", ans[i]+1);
        }
    }
    else
        printf("NO\n");
    return 0;
}