You're given an array a of length 2n. Is it possible to reorder it in such way so that the sum of the first n elements isn't equal to the sum of the last n elements?

Input

The first line contains an integer n (1 ≤ n ≤ 1000), where 2n is the number of elements in the array a.

The second line contains 2n space-separated integers a_1, a_2, …, a_{2n} (1 ≤ a_i ≤ 10^6) — the elements of the array a.

Output

If there's no solution, print "-1" (without quotes). Otherwise, print a single line containing 2n space-separated integers. They must form a reordering of a. You are allowed to not change the order.

Examples

Input


3
1 2 2 1 3 1


Output


2 1 3 1 1 2

Input


1
1 1


Output


-1

Note

In the first example, the first n elements have sum 2+1+3=6 while the last n elements have sum 1+1+2=4. The sums aren't equal.

In the second example, there's no solution.
t=int(input())
l=list(map(int,input().split()))
p=list(l)
p.sort(reverse=True)
my_set=[]
use=[0]*2*t
for i in range(t):
	my_set.append(p[i])
for i in range(t):
	e=my_set[i]
	for j in range(2*t):
		if(e==l[j]):
			if(use[j]==1):
				continue
			else:
				use[j]=1
				break
mp=[]
ps=[]
for i in range(2*t):
	if(use[i]==1):
		mp.append(l[i])

	if(use[i]==0):
		ps.append(l[i])
if(sum(mp)==sum(ps)):
	print(-1)
else:
	print(*(mp+ps))
# s=0
# po=sum(l)
# ans=-1
# my=0
# for i in range(t+1):
# 	s=0
# 	for j in range(i,i+t):
# 		s=s+l[j]
# 	if(s!=po-s):
# 		ans=0
# 		my=i
# 		break
# if(ans==-1):
# 	print(-1)
# else:
# 	op=l[my:my+t]
# 	for i in range(2*t):
# 		if(i<my or i>=my+t):
# 			op.append(l[i])
# 	print(*op)
