A. Diverse Team

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根据学生在Codehorses上的评分，你需要从n个学生中选择k个（1≤k≤n≤100）组成一个团队，要求团队成员的评分各不相同。如果无法形成合适的团队，输出"NO"；否则输出"YES"，并给出k个不同的学生编号，表示所选团队成员。题目提供输入包括学生数量n、团队规模k以及每个学生的评分。输出要求如果存在多个答案，则任意输出一个。

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A. Diverse Team

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are nn students in a school class, the rating of the ii-th student on Codehorses is aiai. You have to form a team consisting of kk students (1≤k≤n1≤k≤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 kk 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 nn and kk (1≤k≤n≤1001≤k≤n≤100) — the number of students and the size of the team you have to form.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100), where aiai is the rating of ii-th student.

Output

If it is impossible to form a suitable team, print "NO" (without quotes). Otherwise print "YES", and then print kk distinct integers from 11 to nnwhich 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 11 to nn.

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.

代码：

#include <iostream>

using namespace std;

int main()
{
    int a[110];
    fill(a,a+110,0);
    int n,m;
    int count=0;
    cin>>n>>m;
    for(int i=1;i<=n;++i)
    {
        int x;
        cin>>x;
        a[x]=i;
    }
     for(int i=1;i<110;++i){
        if(a[i]!=0)
            count++;
     }
     if(count<m) cout<<"NO"<<endl;
     else{
        cout<<"YES"<<endl;
        for(int i=1;i<110;++i){
            if(a[i]!=0&&m!=0)
            {
                cout<<a[i]<<" ";
                m--;
            }
        }
     }
     return 0;
}