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.
//coded by: pranonrahman-------------------------------------
//-----------------------------------------------------------
//-----------------------------------------------------------
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>

#define O_O ios_base::sync_with_stdio(0); cin.tie(NULL)
#define endl '\n'
#define PI acos(-1.0)
#define CASE(n) cout << "Case " << n << ": "
#define CASE_(n) cout << "Case " << n << endl
#define FIXED(n) cout << fixed << setprecision(n)
#define testPrint(n) cout << "test" << n << endl;


int main()
{
    O_O;
    long long t=1;
   // cin >> t;
    long long T = t;
    while(t--)
    {
        //write your code here
        int n;
        cin >> n;
        int arr[n+44];
        int f=1;
        int mxi[n+44];
        map<int,int>mp;
        arr[0]=0;
        for(int i=1;i<=n;i++)
        {
            cin >> arr[i];
            mp[arr[i]]++;
            if(arr[i]>i)    f=0;
            else if(arr[i]<arr[i-1] and i!=0)   f=0;
        }
        int mx=1;
        int b[n+6];
        set<int>st;
        for(int i=0;i<=n;i++)   st.insert(i);
        for(int i=1;i<=n;i++)    st.erase(arr[i]);
        if(f==0)
        {
            cout << -1 << endl;
            return 0;
        }
        for(int i=1;i<=n ;i++)
        {
        //    cout << i << ' ' << *(st.begin()) << ' ' << arr[i] << ' ' << mx << endl;
            b[i]=*(st.begin());
            st.erase(b[i]);
            mp[arr[i]]--;
            if(mp[i]==0)    st.insert(arr[i]);
        }
        for(int i=1;i<=n;i++) cout << b[i] << ' ' ;
        cout << endl;
    }
    return 0;
}



