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.
//
// Created by watemus on 13.06.2020.
//

#ifdef LOCAL
#define _GLIBCXX_DEBUG
#endif

#include <bits/stdc++.h>

using namespace std;

#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff first
#define ss second

using ll = long long;
using ld = long double;

template<typename T>
using vec = vector<T>;

template<typename T>
using uset = unordered_set<T>;

template<typename T1, typename T2>
using umap = unordered_map<T1, T2>;

#define int ll

constexpr ll INFL = 1'000'000'000'000'000'228;
constexpr int INFI = 1'000'000'228;
const ld PI = acos(-1);

vector<pair<int, int>> DD = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

#ifdef LOCAL
#else
#endif

void run() {
    int n;
    cin >> n;
    vec<int> a(n + 1);
    vec<int> b(n + 1);
    set<int> banned;
    vec<int> usd(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        if (a[i] != a[i - 1]) {
            b[i] = a[i - 1];
            usd[i] = 1;
            banned.insert(a[i - 1]);
        }
    }
    int cur = 0;
    for (int i = 1; i <= n; i++) {
        if (usd[i] && b[i] - b[i - 1] > 1) {
            cout << "-1\n";
            return;
        }
        if (!usd[i]) {
            while (banned.count(cur)) {
                cur++;
            }
            b[i] = cur;
            cur++;
        }
    }
    for (int i = 1; i <= n;i++) {
        cout << b[i] << ' ';
    }
    cout << '\n';
}
/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/

signed main() {
#ifdef LOCAL
    std::freopen("input.txt", "r", stdin);
#else
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
#endif
    cout << fixed << setprecision(20);
    int t = 1;
//    cin >> t;
    while (t--) {
        run();
    }
    return 0;
}

