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>
#define ll long long int
#define vpii vector< pair<int,int> >
#define vpll vector< pair<ll,ll> >
#define mp(a,b) make_pair(a,b)
#define mpii map<int,int>
#define mpll map<ll,ll>
#define MOD 1000000007
#define all(v) v.begin(),v.end()
#define s(v) v.size()
#define test ll t;cin>>t;while(t--)
#define vec vector<ll>
#define  pb(a) push_back(a)
using namespace std;
int main()
{
    int n;
    cin>>n;
    int a[n],b[n];
    for(int i=0;i<n;i++)
        cin>>a[i];
    bool flag =0;
    if(a[0]>1)
        flag=1;
    if(a[n-1]>n)
        flag=1;
    for(int i=1;i<n;i++)
    {
        if(a[i]-a[i-1]>(i+1))
            flag=1;
    }
    if(flag==1)
    {
        cout<<-1<<endl;
    }
    else
    {
        bool ok[n]={0};
        b[0]=-1;
        ok[0]=0;
        for(int i=1;i<n;i++)
        {
            if(a[i]!=a[i-1])
            {
                b[i]=a[i-1];
                ok[b[i]]=1;
            }
            else
            {
                b[i]=-1;
            }

        }
        int m=0;
        for(int i=0;i<n;i++)
        {
            while(ok[m]==1)
            {
                m++;
            }
            if(b[i]==-1)
            {
                b[i]=m;
            }
        }
        for(int i=0;i<n;i++)
            cout<<b[i]<<' ';


    }
}
