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;

long long int poow(long long int a,long long int b)
{
    long long int p=1000000007;
    long long int res=1;
    while(b)
    {
        if(b&1)
        res=(res*a)%p;
        a=(a*a)%p;
        b=b>>1;
    }
    return res;
}

int main() {
    
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    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;
    }
    for(int i=1;i<n;i++)
    {
        if(a[i]>i+1 || a[i]<a[i-1])
        {
            cout<<-1;
            return 0;
        }
    }
    if(n==1)
    {
        if(a[0]==0)
        cout<<1;
        else
        cout<<0;
        return 0;
    }
    vector<int> v;
    int vis[n+3]={0};
    v.push_back(a[0]);
    vis[a[0]]=1;
    for(int i=1;i<n;i++)
    if(a[i]!=a[i-1])
    v.push_back(a[i]),vis[a[i]]=1;
    int b[n];
    memset(b,-1,sizeof(b));
    int j=v.size()-1;
    for(int i=n-1;i>=0;i--)
    {
        if(a[i]==v[j])
        {
            j--;
            b[i]=v[j];
        }
        if(j==0)
        break;
    }
    j=0;
    
    for(int i=0;i<n;i++)
    {
        while(vis[j])
        j++;
        if(b[i]==-1)
        b[i]=j,vis[j]=1;
        cout<<b[i]<<" ";
    }
	return 0;
}