160_C. Find Pair_34905
You've got another problem dealing with arrays. Let's consider an arbitrary sequence containing n (not necessarily different) integers a1, a2, ..., an. We are interested in all possible pairs of numbers (ai, aj), (1 ≤ i, j ≤ n). In other words, let's consider all n2 pairs of numbers, picked from the given array.

For example, in sequence a = {3, 1, 5} are 9 pairs of numbers: (3, 3), (3, 1), (3, 5), (1, 3), (1, 1), (1, 5), (5, 3), (5, 1), (5, 5).

Let's sort all resulting pairs lexicographically by non-decreasing. Let us remind you that pair (p1, q1) is lexicographically less than pair (p2, q2) only if either p1 < p2, or p1 = p2 and q1 < q2.

Then the sequence, mentioned above, will be sorted like that: (1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)

Let's number all the pair in the sorted list from 1 to n2. Your task is formulated like this: you should find the k-th pair in the ordered list of all possible pairs of the array you've been given.

Input

The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ n2). The second line contains the array containing n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109). The numbers in the array can coincide. All numbers are separated with spaces.

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout, streams or the %I64d specificator instead.

Output

In the single line print two numbers — the sought k-th pair.

Examples

Input

2 4
2 1


Output

2 2


Input

3 2
3 1 5


Output

1 3

Note

In the first sample the sorted sequence for the given array looks as: (1, 1), (1, 2), (2, 1), (2, 2). The 4-th of them is pair (2, 2).

The sorted sequence for the array from the second sample is given in the statement. The 2-nd pair there is (1, 3).
#import math
#from functools import lru_cache
#import heapq
#from collections import defaultdict
from collections import Counter
#from collections import deque
#from sys import stdout
#from sys import setrecursionlimit
#setrecursionlimit(10**7)
#from bisect import bisect_left
from sys import stdin
input = stdin.readline

INF = 10**9 + 7
MAX = 10**7 + 7
MOD = 10**9 + 7

n, k = [int(x) for x in input().strip().split()]
a = [int(x) for x in input().strip().split()]
c = list(Counter(a).items())
c.sort()
#c.append((0, 0))
s = 0
fi = 0
i = 0
while(i<len(c)):
    s += c[i][1]*n
    #print(i, s)
    if(s>=k):
        fi = i
        k -= (s - c[i][1]*n)
        break
    i+=1
si = 0
i = 0
s = 0
while(i<len(c)):
    s += c[i][1]*c[fi][1]
    #print(i, s)
    if(s>=k):
        si = i
        break
    i+=1
print(c[fi][0], c[si][0])
