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
#define pb push_back
#define all(a) a.begin(),a.end()
#define f(i,l,u) for(ll i=l;i<u;i++)
int main() {
    
     ll n;
     cin>>n;
     ll a[n];
     set<ll> s;
     map<ll,ll> count;
     ll mex=-1;
     ll fl=0;
     vector<ll> v;
     f(i,0,n)
     {
         cin>>a[i];
            mex=max(mex,a[i]);
            if(!count[a[i]])
            v.pb(a[i]);

         count[a[i]]++;
         
     }
     
     
     f(i,0,n)
     if(a[i]>i+1)
     {
         fl=1;
         break;
     }
     
     if(fl)
     cout<<-1<<endl;
     else
     {
         vector<ll> kk;
     ll c=0;
     ll vis[mex]={0};
     f(i,0,mex)
     {
         
         if(!count[i])
      {     kk.pb(i);
          c++;
      
       
      }  
         
     }
     
    // cout<<"C= "<<c<<endl;
    
     ll k=0;
    ll val=1;
    
    
      cout<<1-a[0]<<" ";
        
        if(kk.size()==0)
        val++;
        else
        k++;
             
      f(i,1,n)
      {
          if(a[i]!=a[i-1])
          cout<<a[i-1]<<" ";
          else
          {
              if(k<kk.size())
           {    
               cout<<kk[k]<<" ";
                k++;
               
           }
              else
              {cout<<mex+val<<" ";
              val++;
              }
             
          }
          
     }
      
      
      
         
     }

    return 0;
}