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 ll long long int
#define fast ios_base::sync_with_stdio(false);  cin.tie(NULL); cout.tie(NULL);
#define arsc(a,n) for(long int index=0;index<n;index++) cin>>a[index];
#define arpr(a,n) for(long int index=0;index<n;index++) cout<<a[index]<<" ";cout<<endl;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define ss second
#define ff first
 


int main() {
    fast;
     #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
   ll  t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;
        ll a[n],b[n],i,x;

        for(i=0;i<n;i++){
            cin>>a[i];
        }
        for(i=1;i<n;i++){
            if(a[i]-a[i-1]>2 || a[i]>i+1) {
                cout<<-1<<endl;
                exit(0);
            }

        }
        ll c[n+5]={},j;
        for(i=0;i<n;i++){
           if(a[i]==0) b[i]=10000005;
           else if(i==0 && a[i]==1) b[i]=0;

           else if(a[i]-a[i-1]<=1) b[i]=a[i]-1;
           else {
            b[i]=a[i]-2;
            for(j=1;j<i;j++){
            if(b[j]==b[j-1] && c[j-1]==0 && c[j]==0){
                b[j]=a[i]-1;
                c[j-1]++;
                c[j]++;
                break;
            }
           }
          }
            
        }
        for(i=0;i<n;i++) cout<<b[i]<<" ";
            cout<<endl;
        

       
    }
    return 0;
        
}