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>
#include <bits/stdc++.h>

using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef int tp;
typedef tree<tp,null_type,less<tp>,rb_tree_tag,tree_order_statistics_node_update> oSet;
#define frt(i, s, t) for (int i = s; i <= t; i++)
#define fr(i, s, t) for (int i = s; i < t; i++)
#define frr(i, s, t) for (int i = s; i >= t; i--)
#define srt(a) sort(a.begin(),a.end());
#define mcp(a,b) memcpy ( a, b, b.size());
#define ll long long
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define lb lower_bound
#define ub upper_bound
#define all(c) (c).begin(), (c).end()

int solve()
{
    int n;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++)
    {
       cin>>a[i];
    }
    if(a[0]>1)
    {
       cout<<-1;
       return 0;
    }

    int s=1-a[0];
    int v=0;
    vector<int> f(n,-1);
    int j=0;
    if(a[0]==1)
    {
       v=1;
       f[0]=0;
       j++;
    }
    for(int i=1;i<n;i++)
    {
       if(v==a[i])
       {
          s++;
          continue;
       }
       else
       {
          s=s+(v-a[i]+1);
          if(s<0)
          {
             cout<<-1<<endl;
             return 0;
          }
          int x=a[i]-1-v;
          int k=0;
          while(k<x)
          {
             if(f[i]!=-1)
             {
                k++;
                continue;
             }

             f[j]=v+1;
             j++;
             v=v+1;
             k++;
          }
          v=a[i];
          f[i]=a[i-1];
       }

    }
    for(int i=0;i<n;i++)
    {
       if(f[i]==-1)
       {
          v++;
          f[i]=v;
       }
       cout<<f[i]<<" ";
    }
    return 0;
}
int main()
{
   ios_base::sync_with_stdio(false);cin.tie(NULL);
   //int t;
   int t=1;
   //cin>>t;
   while(t--)
   {
      solve();
   }

   //cerr<<"time taken : "<<(float)clock()/CLOCKS_PER_SEC<<" secs"<<endl;
   return 0;
}
