Given an array a of length n, find another array, b, of length n such that:

  * for each i (1 ≤ i ≤ n) MEX(\\{b_1, b_2, …, b_i\})=a_i. 



The MEX of a set of integers is the smallest non-negative integer that doesn't belong to this set.

If such array doesn't exist, determine this.

Input

The first line contains an integer n (1 ≤ n ≤ 10^5) — the length of the array a.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ i) — the elements of the array a. It's guaranteed that a_i ≤ a_{i+1} for 1≤ i < n.

Output

If there's no such array, print a single line containing -1.

Otherwise, print a single line containing n integers b_1, b_2, …, b_n (0 ≤ b_i ≤ 10^6)

If there are multiple answers, print any.

Examples

Input


3
1 2 3


Output


0 1 2 

Input


4
0 0 0 2


Output


1 3 4 0 

Input


3
1 1 3


Output


0 2 1 

Note

In the second test case, other answers like [1,1,1,0], for example, are valid.
def solve():
   n = int(input())
   lst = list(map(int,input().split()))
   svb = []
   svbc = 0
   ans = []
   pos = 0
   for i in range(n):
       if lst[i] == pos:
           svbc += 1
           ans.append(-1)
       elif lst[i] == pos + 1:
           ans.append(pos)
           pos += 1
       elif lst[i] > pos:
           ans.append(pos)
           for x in range(pos+1, lst[i]):
               svb.append(x)
               svbc -= 1
           pos = lst[i]
           if svbc < 0:
               print(-1)
               return
   sh = -1
   for i in ans:
       if i != -1:
           print(i,end=' ')
       else:
           sh += 1
           if sh >= len(svb):
               print(1000000, end=' ')
           else:
               print(svb[sh], end=' ')
   print()
for t in range(1):
    solve()