Problem: There are n distinct points on a coordinate line, the coordinate of i-th point equals to x_i. Choose a subset of the given set of points such that the distance between each pair of points in a subset is an integral power of two. It is necessary to consider each pair of points, not only adjacent. Note that any subset containing one element satisfies the condition above. Among all these subsets, choose a subset with maximum possible size.

In other words, you have to choose the maximum possible number of points x_{i_1}, x_{i_2}, ..., x_{i_m} such that for each pair x_{i_j}, x_{i_k} it is true that |x_{i_j} - x_{i_k}| = 2^d where d is some non-negative integer number (not necessarily the same for each pair of points).

Input

The first line contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of points.

The second line contains n pairwise distinct integers x_1, x_2, ..., x_n (-10^9 ≤ x_i ≤ 10^9) — the coordinates of points.

Output

In the first line print m — the maximum possible number of points in a subset that satisfies the conditions described above.

In the second line print m integers — the coordinates of points in the subset you have chosen.

If there are multiple answers, print any of them.

Examples

Input

6
3 5 4 7 10 12


Output

3
7 3 5

Input

5
-1 2 5 8 11


Output

1
8

Note

In the first example the answer is [7, 3, 5]. Note, that |7-3|=4=2^2, |7-5|=2=2^1 and |3-5|=2=2^1. You can't find a subset having more points satisfying the required property.

Solution (in PYTHON3):

Solution:
import sys


def solve(io):
	N = io.readInt()
	X = io.readIntArray(N)

	has = set(X)

	best = (X[0],)
	for v in X:
		for b in range(0, 31):
			p = 1 << b
			a = v + p
			if a in has:
				if len(best) < 2:
					best = (v, a)
				b = a + p
				if b in has and len(best) < 3:
					best = (v, a, b)
					break
	io.println(len(best))
	io.printlnArray(best)


# +---------------------+
# | TEMPLATE CODE BELOW |
# |    DO NOT MODIFY    |
# +---------------------+


# TODO: maybe reading byte-by-byte is faster than reading and parsing tokens.
class IO:
	input = None
	output = None

	def __init__(self, inputStream, outputStream):
		self.input = inputStream
		self.output = outputStream

	@staticmethod
	def isSpaceChar(c):
		return c == ' ' or c == '\n' or c == '\0' or c == '\r' or c == '\t'

	@staticmethod
	def isLineBreakChar(c):
		return c == '\n' or c == '\0' or c == '\r'

	def readChar(self):
		return self.input.read(1)

	def readString(self):
		c = self.readChar()
		while IO.isSpaceChar(c):
			c = self.readChar()
		sb = []
		while not IO.isSpaceChar(c):
			sb.append(c)
			c = self.readChar()
		return ''.join(sb)

	def readLine(self):
		c = self.readChar()
		while IO.isSpaceChar(c):
			c = self.readChar()
		sb = []
		while not IO.isLineBreakChar(c):
			sb.append(c)
			c = self.readChar()
		return ''.join(sb)

	def readInt(self):
		c = self.readChar()
		while IO.isSpaceChar(c):
			c = self.readChar()
		sgn = 1
		if c == '-':
			sgn = -1
			c = self.readChar()
		res = 0
		while not IO.isSpaceChar(c):
			res = (res * 10) + ord(c) - 48
			c = self.readChar()
		if sgn < 0:
			return -res
		return res

	def readFloat(self):
		return float(self.readString())

	def readStringArray(self, N, offset=0):
		arr = [None] * offset
		for _ in range(0, N):
			arr.append(self.readString())
		return arr

	def readIntArray(self, N, offset=0):
		arr = [None] * offset
		for _ in range(0, N):
			arr.append(self.readInt())
		return arr

	def readFloatArray(self, N, offset=0):
		arr = [None] * offset
		for _ in range(0, N):
			arr.append(self.readFloat())
		return arr

	def print(self, s):
		self.output.write(str(s))

	def println(self, s):
		self.print(s)
		self.print('\n')
	
	def printlnArray(self, arr, sep = ' '):
		self.println(sep.join(str(x) for x in arr))

	def flushOutput(self):
		self.output.flush()


pythonIO = IO(sys.stdin, sys.stdout)
solve(pythonIO)
pythonIO.flushOutput()