Problem: Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.

For example, the following matrices are palindromic:

<image>

The following matrices are not palindromic because they change after the order of rows is reversed:

<image>

The following matrices are not palindromic because they change after the order of columns is reversed:

<image>

You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".

Input

The first line contains one integer n (1 ≤ n ≤ 20).

The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.

Output

If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.

If it's impossible to construct any matrix, then print "NO".

You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.

Examples

Input


4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1


Output


YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1


Input


3
1 1 1 1 1 3 3 3 3


Output


YES
1 3 1
3 1 3
1 3 1


Input


4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1


Output


NO


Input


1
10


Output


YES
10 

Note

Note that there exist multiple answers for the first two examples.

Solution (in PYTHON3):

Solution:
import io, sys, atexit, os

import math as ma
from sys import exit
from decimal import Decimal as dec
from itertools import permutations


def li ():
	return list (map (int, input ().split ()))


def num ():
	return map (int, input ().split ())


def nu ():
	return int (input ())


def find_gcd ( x, y ):
	while (y):
		x, y = y, x % y
	return x


def solve ():
	t = 1
	for it in range (t):
		n = nu ()
		a = li ()
		ff = [ 0 ] * 1001
		qq = -1
		for i in a:
			ff [ i ] += 1
		zz = 0
		for i in range (len (ff)):
			if (ff [ i ] % 2 == 1):
				zz += 1
				qq = i
		py = -1
		for i in range (n):
			if (a [ i ] == qq):
				py = i
		if (n % 2 == 0):
			if (zz > 0):
				print ("NO")
			else:
				xx = [ 0 ] * n
				for i in range (n):
					xx [ i ] = [ 0 ] * n
				cc = 0
				fl=True
				for i in range (n // 2):
					for j in range (n // 2):
						zp=False
						for k in range (len (ff)):
							if (ff [ k ] >= 4):
								xx [ i ] [ j ] = k
								zp=True
								xx [ i ] [ n - 1 - j ] = k
								xx [ n - 1 - i ] [ j ] = k
								xx [ n - 1 - i ] [ n - 1 - j ] = k
								ff [ k ] -= 4
								break
						if(zp==False):
							fl=False
				if(fl==False):
					print("NO")
					continue
				print ("YES")
				for i in range (n):
					for j in range (n):
						print (xx [ i ] [ j ], end = " ")
					print ()
		else:
			if (zz > 1):
				print ("NO")
			else:
				fl=True
				if(zz==0):
					print("NO")
					continue
				xx = [ 0 ] * n
				for i in range (n):
					xx [ i ] = [ 0 ] * n
				rr = a [ py ]
				xx [ n // 2 ] [ n // 2 ] = qq
				ff [ qq ] -= 1
				cc = 0
				for i in range (n // 2):
					for j in range (n // 2):
						zp=False
						for k in range (len (ff)):
							if (ff [ k ] >= 4):
								xx [ i ] [ j ] = k
								xx [ i ] [ n - 1 - j ] = k
								xx [ n - 1 - i ] [ j ] = k
								xx [ n - 1 - i ] [ n - 1 - j ] = k
								ff [ k ] -= 4
								zp=True
								break
						if(zp==False):
							fl=False
				for i in range (n // 2):
					zp = False
					for j in range (len (ff)):
						if (ff [ j ] >= 2):
							xx [ n // 2 ] [ i ] = j
							xx [ n // 2 ] [ n - 1 - i ] = j
							zp=True
							ff [ j ] -= 2
							break
					if (zp == False):
						fl = False
				for i in range (n // 2):
					zp=True
					for j in range (len (ff)):
						if (ff [ j ] >= 2):
							xx [ i ] [ n // 2 ] = j
							zp=True
							xx [ n-1-i ] [ n // 2 ] = j
							ff [ j ] -= 2
							break
					if (zp == False):
						fl = False
				if(fl==False):
					print("NO")
					continue
				print ("YES")
				for i in range (n):
					for j in range (n):
						print (xx [ i ] [ j ], end = " ")
					print ()


if __name__ == "__main__":
	solve ()