Xenia the mathematician has a sequence consisting of n (n is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three a, b, c the following conditions held:

  * a < b < c; 
  * a divides b, b divides c. 



Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has <image> groups of three.

Help Xenia, find the required partition or else say that it doesn't exist.

Input

The first line contains integer n (3 ≤ n ≤ 99999) — the number of elements in the sequence. The next line contains n positive integers, each of them is at most 7.

It is guaranteed that n is divisible by 3.

Output

If the required partition exists, print <image> groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.

If there is no solution, print -1.

Examples

Input

6
1 1 1 2 2 2


Output

-1


Input

6
2 2 1 1 4 6


Output

1 2 4
1 2 6
"""
 Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
 
def ceil(x):
    if x != int(x):
        x = int(x) + 1;
    return x;
 
def factorial(x, m):
	val = 1
	while x>0:
		val = (val * x) % m
		x -= 1
	return val
    
# swap_array function
def swaparr(arr, a,b):
    temp = arr[a];
    arr[a] = arr[b];
    arr[b] = temp;
 
## gcd function
def gcd(a,b):
    if b == 0:
        return a;
    return gcd(b, a % b);
 
## nCr function efficient using Binomial Cofficient
def nCr(n, k): 
    if(k > n - k): 
        k = n - k; 
    res = 1;
    for i in range(k): 
        res = res * (n - i);
        res = res / (i + 1); 
    return int(res);
 
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
    if hi == None:
        hi = len(a);
    while lo < hi:
        mid = (lo+hi)//2;
        if a[mid] < x:
            lo = mid+1;
        else:
            hi = mid;
    return lo;
 
## prime factorization
def primefs(n):
    ## if n == 1    ## calculating primes
    primes = {}
    while(n%2 == 0 and n > 0):
        primes[2] = primes.get(2, 0) + 1
        n = n//2
    for i in range(3, int(n**0.5)+2, 2):
        while(n%i == 0 and n > 0):
            primes[i] = primes.get(i, 0) + 1
            n = n//i
    if n > 2:
        primes[n] = primes.get(n, 0) + 1
    ## prime factoriazation of n is stored in dictionary
    ## primes and can be accesed. O(sqrt n)
    return primes
 
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p): 
    res = 1
    x = x % p  
    if (x == 0) : 
        return 0
    while (y > 0) : 
        if ((y & 1) == 1) : 
            res = (res * x) % p 
        y = y >> 1      
        x = (x * x) % p 
    return res 
 
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
    temp = a
    a = b
    b = temp
    return a,b;
 
# find function with path compression included (recursive)
# def find(x, link):
#     if link[x] == x:
#         return x
#     link[x] = find(link[x], link);
#     return link[x];
 
# find function with path compression (ITERATIVE)
def find(x, link):
    p = x;
    while( p != link[p]):
        p = link[p];
    
    while( x != p):
        nex = link[x];
        link[x] = p;
        x = nex;
    return p;
 
 
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
    x = find(x, link)
    y = find(y, link)
    if size[x] < size[y]:
        x,y = swap(x,y)
    if x != y:
        size[x] += size[y]
        link[y] = x
 
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n): 
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    return prime
 
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
    spf[1] = 1;
    for i in range(2, MAXN):
        spf[i] = i;
    for i in range(4, MAXN, 2):
        spf[i] = 2;
    for i in range(3, ceil(MAXN ** 0.5), 2):
        if spf[i] == i:
            for j in range(i*i, MAXN, i):
                if spf[j] == j:
                    spf[j] = i;
    ## function for storing smallest prime factors (spf) in the array
 
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
    ret = {};
    while x != 1:
        ret[spf[x]] = ret.get(spf[x], 0) + 1;
        x = x//spf[x]
    return ret;
    ## this function is useful for multiple queries only, o/w use
    ## primefs function above. complexity O(log n)
 
## taking integer array input
def int_array():
    return list(map(int, input().strip().split()));
 
def float_array():
    return list(map(float, input().strip().split()));
 
## taking string array input
def str_array():
    return input().strip().split();
 
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
 
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
 
from itertools import permutations
 
def solve():
	n = int(input())
	a = list(map(int, input().split()))
	ref = [[1, 2, 4], [1, 2, 6], [1, 3, 6]]
	d = {}
	one = 0
	two = 0
	three = 0
	four = 0
	six = 0
	for x in a:
		if x == 1:
			one += 1
		elif x == 2:
			two += 1
		elif x == 3:
			three += 1
		elif x == 4:
			four += 1
		elif x == 6:
			six += 1
 
	if one == n // 3:
		if two + three == n // 3 and four + six == n // 3:
			## we have to check whether three don't get aligned with the four ##
			if three > six:
				flag = 0
			else:
				flag = 1
		else:
			flag = 0
	else:
		flag = 0
 
	if flag == 1:
		for i in range(n//3):
			if one > 0 and two > 0 and four > 0:
				print(1, 2, 4)
				one, two, four = (one - 1, two - 1, four - 1)
			elif one > 0 and two > 0 and six > 0:
				print(1, 2, 6)
				one, two , six = (one - 1, two - 1, six - 1)
			elif one > 0 and three > 0 and six > 0:
				print(1, 3, 6)
				one, three, six = (one - 1, three - 1, six - 1)
	else:
		print(-1)
 
			
 
if __name__ == '__main__':
	for _ in range(1):
		solve()
	# fin_time = datetime.now()
# 	print("Execution time (for loop): ", (fin_time-init_time))
 