SDUT 算法训练赛 contest 5(set,vis共用)

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 nn 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 11 to nn.

Examples

Input

5 3
15 13 15 15 12

Output

YES
1 2 5 

Input

5 4
15 13 15 15 12

Output

NO

Input

4 4
20 10 40 30

Output

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.

大佬：

#include<bits/stdc++.h>
using namespace std;
set<int>se;
int vis[110];
int main()
{
    int n,m,a[110];
    cin>>n>>m;
    for(int i=1; i<=n; i++)
        cin>>a[i],se.insert(a[i]);
    if(se.size()<m) puts("NO");
    else
    {
        puts("YES");
   int cnt=0;
        for(int i=1; i<=n; i++)
            if(!vis[a[i]]&&cnt<m)
                vis[a[i]]=1,cnt++,cout<<i<<" ";
    }
    return 0;
}

MINE:

#include<iostream>
using namespace std;
int tong[105]= {0};
int main()
{
    int a[105];
    int n,k,sum=0;
    cin>>n>>k;
    for(int i=0; i<n; i++)
    {
        cin>>a[i];
        tong[a[i]]=1;
    }
    for(int j=1; j<=100; j++)
        if(tong[j]==1)
            sum++;
    if(sum<k)
        cout<<"NO"<<endl;
    else
    {
        cout<<"YES"<<endl;
        for(int i=0; i<n; i++)
            if(tong[a[i]]==1)
            {
                if(k==0)
                    break;
                k--;
                if(k==0)
                cout<<i+1<<endl;
                else
                cout<<i+1<<" ";
                tong[a[i]]=0;
            }
    }
    return 0;
}