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:

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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:

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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:
from collections import Counter

def solve(n, nums):
    if n == 1:
        return False if len(nums) < 1 else [[nums.pop()]]

    counts = Counter(nums)
    if n % 2 == 0:
        keys = []
        for num, count in counts.items():
            if count >= 4 and count % 4 == 0:
                while count > 0:
                    keys.append(num)
                    count -= 4
            else:
                return False
        
        matrix = [None for _ in range(n)]

        for r in range(0, n//2):
            row = [0] * n
            for c in range(0, n//2):
                key = keys.pop()
                row[c] = key
                row[n-c-1] = key
            matrix[r] = row
            matrix[n-r-1] = row
        
        return matrix
    else:
        num_fours_needed = (n//2) * (n//2)
        keys = []
        for key, count in counts.items():
            while count >= 4 and len(keys) < num_fours_needed:
                keys.append(key)
                count -= 4
            
            counts[key] = count

            if len(keys) == num_fours_needed:
                break
        
        if len(keys) != num_fours_needed:
            return False

        count_nums = {1: [], 2:[]}

        for key, count in counts.items():
            while count >= 4:
                count_nums[2].append(key)
                count -= 2
            
            while count >= 2:
                count_nums[2].append(key)
                count -= 2

            if count != 0:
                count_nums[count].append(key)

        matrix = [[] for _ in range(n)]
        #print(keys)
        #print(num_fours_needed)
        for r in range(0, n//2):
            row = [0] * n
            for c in range(0, n//2):
                key = keys.pop()
                row[c] = key
                row[n-c-1] = key
            matrix[r] = row
            matrix[n-r-1] = row
        
        try:
            # Fill in middle row
            row = [0] * n
            matrix[n//2] = row
            #print(matrix)

            for i in range(0, n//2):
                key = count_nums[2].pop()
                row[i] = key
                row[n-i-1] = key
            
            row[n//2] = count_nums[1].pop()

            # Fill in middle column
            mid_col = n//2
            for i in range(0, n//2):
                key = count_nums[2].pop()
                matrix[i][mid_col] = key
                matrix[n-i-1][mid_col] = key
        except Exception as E:
            return False

        return matrix
        


        


            
n = int(input())
nums = list(map(int, input().split()))

res = solve(n, nums)

if res:
    print("YES")
    for r in res:
        print(*map(str, r))
else:
    print("NO")