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C. Find Pair

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputYou've got another problem dealing with arrays. Let's consider an arbitrary sequence containing *n* (not necessarily different) integers *a*_{1}, *a*_{2}, ..., *a*_{n}. We are interested in all possible pairs of numbers (*a*_{i}, *a*_{j}), (1 ≤ *i*, *j* ≤ *n*). In other words, let's consider all *n*^{2} 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 (*p*_{1}, *q*_{1}) is lexicographically less than pair (*p*_{2}, *q*_{2}) only if either *p*_{1} < *p*_{2}, or *p*_{1} = *p*_{2} and *q*_{1} < *q*_{2}.

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 *n*^{2}. 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* ≤ 10^{5}, 1 ≤ *k* ≤ *n*^{2}). The second line contains the array containing *n* integers *a*_{1}, *a*_{2}, ..., *a*_{n} ( - 10^{9} ≤ *a*_{i} ≤ 10^{9}). 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).

Codeforces (c) Copyright 2010-2019 Mike Mirzayanov

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