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.
#include <bits/stdc++.h>
using namespace std;

#define int long long int
#define mod 1000000007
#define f0(n) for(int i=0;i<n;i++)
#define vi vector<int>
#define vs vector<string>
#define mp make_pair
#define pb push_back
#define rt return
#define sumof(arr) accumulate(arr,arr+n,0);
double pi = 3.141592653589793238462;
string numkey="0123456789";
string uni="abcdefghijklmnopqrstuvwxyz";
#define pii pair<int,int>
#define setf cout.setf(ios::fixed);
#define setp setprecision
#define mii map<int,int>
#define F first
#define S second
#define all(x) x.begin(),x.end()

int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};

int INF=(1LL<<62);


signed main()
{
  #ifndef ONLINE_JUDGE
  freopen("input.txt","r",stdin);
  freopen("out1.txt","w",stdout);
  #endif
  
  int n;
  cin>>n;
  int arr[n+1]={0};
  int brr[n+1];
  memset(brr,-1,sizeof(brr));
  for(int i=1;i<=n;i++)
  	cin>>arr[i];

  set<int>s;

  for(int i=1;i<=n;i++)
  {
  	if(arr[i-1]!=arr[i])
  	{
  		brr[i]=arr[i-1];
  		s.insert(arr[i-1]);

  	}

  }
  s.insert(arr[n]);
  int m=0;
  for(int i=1;i<=m;i++)
  {
  	while(s.count(m))
  		m++;
  	if(brr[i]==-1)
  	{
  		brr[i]=m;
  		s.insert(m);

  	}

  }
  for(int i=1;i<=n;i++)
  	cout<<brr[i]<<" ";




    
  return 0;
}