Problem: You are given an array s consisting of n integers.

You have to find any array t of length k such that you can cut out maximum number of copies of array t from array s.

Cutting out the copy of t means that for each element t_i of array t you have to find t_i in s and remove it from s. If for some t_i you cannot find such element in s, then you cannot cut out one more copy of t. The both arrays can contain duplicate elements.

For example, if s = [1, 2, 3, 2, 4, 3, 1] and k = 3 then one of the possible answers is t = [1, 2, 3]. This array t can be cut out 2 times. 

  * To cut out the first copy of t you can use the elements [1, \underline{2}, 3, 2, 4, \underline{3}, \underline{1}] (use the highlighted elements). After cutting out the first copy of t the array s can look like [1, 3, 2, 4]. 
  * To cut out the second copy of t you can use the elements [\underline{1}, \underline{3}, \underline{2}, 4]. After cutting out the second copy of t the array s will be [4]. 



Your task is to find such array t that you can cut out the copy of t from s maximum number of times. If there are multiple answers, you may choose any of them.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in s and the desired number of elements in t, respectively.

The second line of the input contains exactly n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 2 ⋅ 10^5).

Output

Print k integers — the elements of array t such that you can cut out maximum possible number of copies of this array from s. If there are multiple answers, print any of them. The required array t can contain duplicate elements. All the elements of t (t_1, t_2, ..., t_k) should satisfy the following condition: 1 ≤ t_i ≤ 2 ⋅ 10^5.

Examples

Input


7 3
1 2 3 2 4 3 1


Output


1 2 3 


Input


10 4
1 3 1 3 10 3 7 7 12 3


Output


7 3 1 3


Input


15 2
1 2 1 1 1 2 1 1 2 1 2 1 1 1 1


Output


1 1 

Note

The first example is described in the problem statement.

In the second example the only answer is [7, 3, 1, 3] and any its permutations. It can be shown that you cannot choose any other array such that the maximum number of copies you can cut out would be equal to 2.

In the third example the array t can be cut out 5 times.

Solution (in PYTHON3):

Solution:
''' Thruth can only be found at one place - THE CODE '''

''' Copyright 2018, SATYAM KUMAR'''
from sys import stdin, stdout
import cProfile, math
from collections import Counter
from bisect import bisect_left,bisect,bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
sys.setrecursionlimit(10**6) # max depth of recursion
threading.stack_size(2**27)  # new thread will get stack of such size

printHeap = str()
memory_constrained = False
P = 10**9+7
import sys
sys.setrecursionlimit(10000000)

class Operation:
    def __init__(self, name, function, function_on_equal, neutral_value=0):
        self.name = name
        self.f = function
        self.f_on_equal = function_on_equal
def add_multiple(x, count):
    return x * count
def min_multiple(x, count):
    return x
def max_multiple(x, count):
    return x
sum_operation = Operation("sum", sum, add_multiple, 0)
min_operation = Operation("min", min, min_multiple, 1e9)
max_operation = Operation("max", max, max_multiple, -1e9)
class SegmentTree:
    def __init__(self,
                 array,
                 operations=[sum_operation, min_operation, max_operation]):
        self.array = array
        if type(operations) != list:
            raise TypeError("operations must be a list")
        self.operations = {}
        for op in operations:
            self.operations[op.name] = op
        self.root = SegmentTreeNode(0, len(array) - 1, self)
    def query(self, start, end, operation_name):
        if self.operations.get(operation_name) == None:
            raise Exception("This operation is not available")
        return self.root._query(start, end, self.operations[operation_name])
    def summary(self):
        return self.root.values
    def update(self, position, value):
        self.root._update(position, value)
    def update_range(self, start, end, value):
        self.root._update_range(start, end, value)
    def __repr__(self):
        return self.root.__repr__()
class SegmentTreeNode:
    def __init__(self, start, end, segment_tree):
        self.range = (start, end)
        self.parent_tree = segment_tree
        self.range_value = None
        self.values = {}
        self.left = None
        self.right = None
        if start == end:
            self._sync()
            return
        self.left = SegmentTreeNode(start, start + (end - start) // 2,
                                    segment_tree)
        self.right = SegmentTreeNode(start + (end - start) // 2 + 1, end,
                                     segment_tree)
        self._sync()
    def _query(self, start, end, operation):
        if end < self.range[0] or start > self.range[1]:
            return None
        if start <= self.range[0] and self.range[1] <= end:
            return self.values[operation.name]
        self._push()
        left_res = self.left._query(start, end,
                                    operation) if self.left else None
        right_res = self.right._query(start, end,
                                      operation) if self.right else None
        if left_res is None:
            return right_res
        if right_res is None:
            return left_res
        return operation.f([left_res, right_res])
    def _update(self, position, value):
        if position < self.range[0] or position > self.range[1]:
            return
        if position == self.range[0] and self.range[1] == position:
            self.parent_tree.array[position] = value
            self._sync()
            return
        self._push()
        self.left._update(position, value)
        self.right._update(position, value)
        self._sync()
    def _update_range(self, start, end, value):
        if end < self.range[0] or start > self.range[1]:
            return
        if start <= self.range[0] and self.range[1] <= end:
            self.range_value = value
            self._sync()
            return
        self._push()
        self.left._update_range(start, end, value)
        self.right._update_range(start, end, value)
        self._sync()
    def _sync(self):
        if self.range[0] == self.range[1]:
            for op in self.parent_tree.operations.values():
                current_value = self.parent_tree.array[self.range[0]]
                if self.range_value is not None:
                    current_value = self.range_value
                self.values[op.name] = op.f([current_value])
        else:
            for op in self.parent_tree.operations.values():
                result = op.f(
                    [self.left.values[op.name], self.right.values[op.name]])
                if self.range_value is not None:
                    bound_length = self.range[1] - self.range[0] + 1
                    result = op.f_on_equal(self.range_value, bound_length)
                self.values[op.name] = result
    def _push(self):
        if self.range_value is None:
            return
        if self.left:
            self.left.range_value = self.range_value
            self.right.range_value = self.range_value
            self.left._sync()
            self.right._sync()
            self.range_value = None
    def __repr__(self):
        ans = "({}, {}): {}\n".format(self.range[0], self.range[1],
                                      self.values)
        if self.left:
            ans += self.left.__repr__()
        if self.right:
            ans += self.right.__repr__()
        return ans

def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")

def primeFactors(n): #n**0.5 complex 
    factors = dict()
    for i in range(2,math.ceil(math.sqrt(n))+1):  
        while n % i== 0: 
            if i in factors:
                factors[i]+=1
            else: factors[i]=1
            n = n // i 
    if n>2:
        factors[n]=1
    return (factors)
    
def isprime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True

def test_print(*args):
    if testingMode:
        print(args)

def display_list(list1, sep=" "):
    stdout.write(sep.join(map(str, list1)) + "\n")

def get_int():
    return int(stdin.readline().strip())

def get_tuple():
    return map(int, stdin.readline().split())

def get_list():
    return list(map(int, stdin.readline().split()))

memory = dict()
def clear_cache():
    global memory
    memory = dict()
def cached_fn(fn, *args):
    global memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result


# -------------------------------------------------------------- MAIN PROGRAM
TestCases = False
testingMode = False
optimiseForReccursion = False #Can not be used clubbed with TestCases
def sortingfunc(n):
    global counts
    return counts[n]    
counts = 0
def main():
    global counts
    n,k = get_tuple()
    li = get_list()
    counts = Counter(li)
    keys = list(counts.keys())
    keys.sort(key=sortingfunc,reverse=True)
    max_possible = sum(counts[keys[i]] for i in range(k) if i<len(keys))//k + 1
    #print(max_possible)
    for i in range(max_possible,0,-1):
        c = 0
        for ki in keys:
            c+=counts[ki]//i
        if c>=k:
            #print(i)
            break
    res = []
    #print(i)
    for j in keys:
        res+=[j]*(counts[j]//i)
        if len(res)>k: break
    #print(res,k)
    display_list(res[:k])




# --------------------------------------------------------------------- END


if TestCases: 
    for _ in range(get_int()): 
        cProfile.run('main()') if testingMode else main() 
else: (cProfile.run('main()') if testingMode else main()) if not optimiseForReccursion else threading.Thread(target=main).start()