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;
void allprime(int n, bool prime[]) {
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
void bfs(vector<int> adj[], int src, bool vis[], int level[]) {
  queue<pair<int, int> > q;
  q.push(make_pair(src, 0));
  vis[src] = true;
  level[src] = 0;
  while (!q.empty()) {
    pair<int, int> vpair = q.front();
    q.pop();
    int v = vpair.first;
    int lvl = vpair.second;
    for (int i = 0; i < adj[v].size(); i++) {
      int next = adj[v][i];
      if (!vis[next]) {
        q.push(make_pair(next, lvl + 1));
        vis[next] = true;
        level[next] = lvl + 1;
      }
    }
  }
}
void dfs(vector<int> adj[], int src, bool vis[]) {
  vis[src] = true;
  for (int i = 0; i < adj[src].size(); i++) {
    int next = adj[src][i];
    if (!vis[next]) {
      dfs(adj, next, vis);
    }
  }
}
void solve() {
  int n;
  cin >> n;
  int a[n], b[n];
  cin >> a[0];
  if (a[0] == 1)
    b[0] = 0;
  else
    b[0] = -1;
  vector<pair<int, int> > v;
  for (int i = 1; i < n; i++) {
    cin >> a[i];
    if (a[i] > a[i - 1]) {
      b[i] = a[i - 1];
      for (int j = a[i - 1] + 1; j < a[i]; j++) v.push_back(make_pair(j, i));
    } else {
      b[i] = -1;
    }
  }
  int j = 0;
  bool flag = true;
  for (int i = 0; i < n; i++) {
    if (b[i] == -1) {
      if (j < v.size()) {
        b[i] = v[j].first;
        if (i >= v[j].second) {
          flag = false;
          break;
        }
        ++j;
      } else
        b[i] = 100001;
    }
  }
  if (flag) {
    for (int i = 0; i < n; i++) cout << b[i] << " ";
    cout << endl;
  } else
    cout << -1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}
