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 vll vector<ll>
#define mll map<ll,ll>
#define pb push_back
#define mp make_pair
#define endl "\n"
#define MAX 1e18
#define F first
#define S second
#define all(s) s.begin(),s.end()
#define mod 998244353
ll lcm(ll a,ll b){return (a*b)/(__gcd(a,b));}
ll cel(ll n,ll d){return (n+d-1)/d;}

void hertz()
{
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    #ifndef ONLINE_JUDGE
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
    #endif
}

int main()
{
    hertz();
    ll n;
    cin>>n;
    ll arr[n];
    mll m;
    for (int i = 0; i < n; ++i)
    {
    	cin>>arr[i];
    	m[arr[i]]++;
    }
    vll ans(n,-1);
    for (int i = 1; i < n; ++i)
    {
    	if(arr[i]!=arr[i-1])
    		ans[i]=arr[i-1];
    }
    ll cur=0;
    for (int i = 0; i < n; ++i)
    {
    	if(ans[i]==-1)
    	{
    		if(m.count(cur)){
	    		while(m[cur])
    				cur++;
    		}
    		ans[i]=cur;
    	}
    }
    for(ll i:ans)
    	cout<<i<<" ";
    cout<<endl;
}