Problem:
Score : 400 points

Problem Statement

Snuke has an empty sequence a.

He will perform N operations on this sequence.

In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).

You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq b_i \leq N

Input

Input is given from Standard Input in the following format:

N
b_1 \dots b_N

Output

If there is no sequence of N operations after which a would be equal to b, print -1.
If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.

Sample Input 1

3
1 2 1

Sample Output 1

1
1
2

In this sequence of operations, the sequence a changes as follows:

After the first operation: (1)

After the second operation: (1,1)

After the third operation: (1,2,1)

Sample Input 2

2
2 2

Sample Output 2

-1

2 cannot be inserted at the beginning of the sequence, so this is impossible.

Sample Input 3

9
1 1 1 2 2 1 2 3 2

Sample Output 3

1
2
2
3
1
2
2
1
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// スタックサイズ: 100MB

#include <iostream>
#include <fstream>
#include <vector>
#include <map>
#include <unordered_map>
#include <string>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <functional>
#include <numeric>
#include <iomanip>
#include <queue>
#include <list>
#include <set>
using namespace std;

typedef int64_t ll;
typedef vector<ll> vll;
typedef pair<ll, ll> pll;
const ll INF = 1000000000000000000;
#define fori(i, a, b) for (ll i = (a); i < (b); ++i)
#define ford(i, a, b) for (ll i = (b - 1); (a) <= i; --i)
#define rep(i, n) fori(i, 0, n)
#define all(v) (v).begin(), (v).end()
ll pw(ll x, ll y) { ll r = 1;  rep(i, y) { r *= x; }; return r; }


int main() {

	ll N;
	cin >> N;
	vll b(N);
	rep(i, N) cin >> b[i];
	
	vll ans;
	while(!b.empty()) {
		ll flag = true;
		ford(i, 1, N + 1) {
			if (i <= b.size() && b[i - 1] == i) {
				ans.push_back(b[i - 1]);
				b.erase(b.begin() + i - 1);
				flag = false;
				break;
			}
		}
		if (flag) {
			cout << -1 << endl;
			return 0;
		}
	}

	reverse(all(ans));
	rep(i, N) cout << ans[i] << endl;
	return 0;
}