Problem: Fillomino is a classic logic puzzle. (You do not need to know Fillomino in order to solve this problem.) In one classroom in Yunqi town, some volunteers are playing a board game variant of it:

Consider an n by n chessboard. Its rows are numbered from 1 to n from the top to the bottom. Its columns are numbered from 1 to n from the left to the right. A cell on an intersection of x-th row and y-th column is denoted (x, y). The main diagonal of the chessboard is cells (x, x) for all 1 ≤ x ≤ n.

A permutation of \{1, 2, 3, ..., n\} is written on the main diagonal of the chessboard. There is exactly one number written on each of the cells. The problem is to partition the cells under and on the main diagonal (there are exactly 1+2+ … +n such cells) into n connected regions satisfying the following constraints:

  1. Every region should be connected. That means that we can move from any cell of a region to any other cell of the same region visiting only cells of the same region and moving from a cell to an adjacent cell. 
  2. The x-th region should contain cell on the main diagonal with number x for all 1≤ x≤ n. 
  3. The number of cells that belong to the x-th region should be equal to x for all 1≤ x≤ n. 
  4. Each cell under and on the main diagonal should belong to exactly one region. 

Input

The first line contains a single integer n (1≤ n ≤ 500) denoting the size of the chessboard.

The second line contains n integers p_1, p_2, ..., p_n. p_i is the number written on cell (i, i). It is guaranteed that each integer from \{1, …, n\} appears exactly once in p_1, ..., p_n.

Output

If no solution exists, output -1.

Otherwise, output n lines. The i-th line should contain i numbers. The j-th number on the i-th line should be x if cell (i, j) belongs to the the region with x cells.

Examples

Input


3
2 3 1


Output


2
2 3
3 3 1


Input


5
1 2 3 4 5


Output


1
2 2
3 3 3
4 4 4 4
5 5 5 5 5

Note

The solutions to the examples are illustrated in the following pictures: <image> <image>

Solution (in PYTHON3):

Solution:
import os,sys;from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno();self.buffer = BytesIO();self.writable = "x" in file.mode or "r" not in file.mode;self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:break
            ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE));self.newlines = b.count(b"\n") + (not b);ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:os.write(self._fd, self.buffer.getvalue());self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file);self.flush = self.buffer.flush;self.writable = self.buffer.writable;self.write = lambda s: self.buffer.write(s.encode("ascii"));self.read = lambda: self.buffer.read().decode("ascii");self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);from collections import *;from math import *
# abc = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
# sys.setrecursionlimit(500000)
###################### Start Here ######################
# from functools import lru_cache
# from collections import defaultdict as dd 
from collections import deque as dq 

n = ii1()
arr = iia()
# first priority down side then left if present
dp = [[None]*n for i in range(n)]
for i in range(n):
    dp[i][i]=arr[i]-1
# jitne value utna chalana hai 
def trav(i,j,rem,val):
    if rem==0:
        return 
    if j-1>=0 and dp[i][j-1]==None:
        dp[i][j-1]=val 
        trav(i,j-1,rem-1,val)
    elif i+1>=0 and dp[i+1][j]==None:
        dp[i+1][j]=val
        trav(i+1,j,rem-1,val)
    else:
        return 


for i in range(n):
    trav(i,i,arr[i]-1,arr[i]-1)
flag = True
for i in range(n):
    for j in range(i+1):
        if dp[i][j]==None:
            flag = False 
        else:
            dp[i][j]+=1
if flag:
    for i in range(n):
        print(*dp[i][:i+1])
else:
    print(-1)