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:
from collections import defaultdict, deque
import sys


def input():
    return sys.stdin.readline().rstrip()


def getcnt(LIST):
    d = defaultdict(int)
    for v in LIST:
        d[v] += 1
    return d


def odd_only_1(dic):
    cnt = 0
    res_value = 0
    for key, value in dic.items():
        if value % 2 != 0:
            cnt += 1
            res_value = key
    if cnt == 1:
        return res_value
    return 0


def slv():
    n = int(input())
    a = list(map(int, input().split()))
    d = getcnt(a)

    if n % 2 == 0:
        #even case
        if all(value % 4 == 0 for value in d.values()):
            stack = deque([])
            for key, value in d.items():
                for _ in range(value):
                    stack.append(key)

            print("YES")
            ans = [[-1]*(n) for i in range(n)]
            for i in range(n//2):
                for j in range(n//2):
                    nums = stack[0]
                    ans[i][j] = ans[i][-(j + 1)] = ans[-(i + 1)
                                                       ][j] = ans[-(i + 1)][-(j + 1)] = nums
                    for _ in range(4):
                        stack.popleft()
            for i in range(n):
                print(*ans[i])
            return
        else:
            print("NO")
        return

    else:
        if not odd_only_1(d):
            print("NO")
            return

        val = odd_only_1(d)
        d[val] -= 1
        ans = [[0]*(n) for i in range(n)]
        ans[n//2][n//2] = val

        if sum(value//4 for value in d.values()) < (n//2) ** 2:
            print("NO")
            return

        print("YES")
        stack = deque([])

        for key, value in d.items():
            u, v = divmod(value, 4)
            for _ in range(4 * u):
                stack.appendleft(key)
            for _ in range(v):
                stack.append(key)

        for i in range(n//2):
            for j in range(n//2):
                nums = stack[0]
                ans[i][j] = ans[i][-(j + 1)] = ans[-(i + 1)
                                                   ][j] = ans[-(i + 1)][-(j + 1)] = nums
                for _ in range(4):
                    stack.popleft()

        for i in range(n//2):
            num = stack[0]
            ans[n//2][i] = ans[n//2][-(i + 1)] = num
            for _ in range(2):
                stack.popleft()

        for i in range(n//2):
            num = stack[0]
            ans[i][n//2] = ans[-(i + 1)][n//2] = num
            for _ in range(2):
                stack.popleft()

        for i in range(n):
            print(*ans[i])

    return


def main():
    t = 1
    for i in range(t):
        slv()
    return


if __name__ == "__main__":
    main()