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 ≤ 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 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 ≤ k ≤ n ≤ 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, ..., a_n (1 ≤ a_i ≤ 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

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.
__author__ = 'Devesh Bajpai'

'''
https://codeforces.com/problemset/problem/988/A

Solution: Using a brute-force approach, where a visited boolean array checks the presence of a number being seen or not.
This insures distinctness constraint. A result list maintains the final indices but indexed at 1. 
As soon as the result list since reaches k, we return the decision YES and result list. If the loop that traverses on the
numbers exhausts and the result list hasn't reached k, the decision NO and null is returned. 
'''


def solve(n, k, nums):
    visited = [False for _ in range(101)]
    result = []
    currResultSize = 0

    for i in xrange(0, n):

        num = nums[i]
        if not visited[num]:

            visited[num] = True
            result.append(i+1) #since the index referred in this problem starts at 1 instead of 0
            currResultSize += 1

            if currResultSize == k:
                return "YES", result

    return "NO", None


if __name__ == "__main__":
    n, k = map(int,raw_input().split(" "))
    nums = map(int,raw_input().split(" "))
    decision, results = solve(n, k, nums)

    print decision
    if decision == "YES":
        print ' '.join([str(result) for result in results])
