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<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
/*
    s.find_by_order(k) ---> returns iterator to the element at index(from 0) k
    s.order_of_key(val) ---> number of elements that are strictly smaller than the given value
*/
typedef long long int ll;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<int> vi;
typedef vector<long long int> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long int>> vvl;
typedef pair<int, int> pii;
typedef pair<long long int, long long int> pll;
typedef vector<pair<int, int>> vii;
typedef vector<pair<long long int, long long int>> vll;
#define fastIO() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define test() long long int tc; cin >> tc; cin.ignore(); while(tc--) 
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define mp make_pair
#define MOD 1000000007
#define SMOD 998244353
#define HASH 18014398241046527
#define pq priority_queue
#define all(x) x.begin(),x.end()
#define SP(x) setprecision(x)
#define sp " "
#define inp(a) ll a;cin>>a;cin.ignore()
#define inps(a,b) ll a,b;cin>>a>>b;cin.ignore()
#define out(a) cout<<SP(20)<<a<<'\n'
#define outs(a,b) cout<<SP(20)<<a<<" "<<b<<'\n'
#define ins insert
#define er erase
#define lb lower_bound
#define ub upper_bound
#define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";
#define inpa(a,n) ll a[n]; for(int i=0; i<n; i++){cin>>a[i];}
#define inpv(a,n) vl a; for(int i=0; i<n; i++){inp(x);a.pb(x);}
#define outa(a,n) for(int i=0;i<n;i++){cout<<a[i]<<" ";}cout<<'\n';
#define outv(v) for(auto i:v){cout<<i<<" ";}cout<<'\n';
set<char> vow = {'a','e','i','o','u','A','E','I','O','U'};
ll xpow(ll n, ll m){if(m==0){return 1;}else{ll x = xpow(n, m/2);if(m&1){return x*x*n;}else{return x*x;}}}
ll xmod(ll a, ll m){return (((a%m)+m)%m);}
ll xmodpow(ll n, ll m, ll y){if(m==0){return 1;}else{ll x = xmodpow(n, m/2, y);if(m&1){return xmod((xmod((xmod(x,y)*xmod(x,y)),y)*xmod(n,y)),y);}else{return xmod((xmod(x,y)*xmod(x,y)),y);}}}

ll dx[] = {1, 0, -1, 0};
ll dy[] = {0, 1, 0, -1};





void SOLVE(){
    ll n;
    cin >> n;
    ll a[n+5] = {0};
    for(ll i=1; i<=n; i++){
        cin >> a[i];
    }
    ll b[n+5] = {0};
    ll ind = 1;
    ll first = 0;
    while(ind<=n){
        ll first2 = 0;
        for(ll i=ind; i<=n; i++){
            if(a[i]!=a[first]){
                first2 = i;
                break;
            }
        }
        if(first2==0){
            for(ll i=ind; i<=n; i++){
                b[i] = 1000000;
            }
            break;
        }
        b[first2] = a[first];
        set <ll> ss;
        for(ll i=a[first]+1; i<=a[first2]-1; i++){
            ss.ins(i);
        }
        for(ll i=ind; i<=first2-1; i++){
            if(ss.size()){
                b[i] = *ss.begin();
                ss.er(b[i]);
            }
            else{
                b[i] = 1000000;
            }
        }
        if(ss.size()){
            cout << -1 << '\n';
            return;
        }
        first = first2;
        ind = first2+1;
    }
    for(ll i=1; i<=n; i++){
        cout << b[i] << ' ';
    }
    cout << '\n';
}

int main() {
	fastIO();
	// #ifndef ONLINE_JUDGE
    //     freopen("i.txt", "r", stdin);
    //     freopen("o.txt", "w", stdout);
	// #endif

    ll TC = 1;
    // cin >> TC;
    for(ll tc1=1; tc1<=TC; tc1++){
        // cout << "Case #" << tc1 << ": ";
        SOLVE();
    }
    printclock;
}