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)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <array>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <stdio.h>
#include <cstdio>
#include <cstring>
#include <math.h>

using namespace std;

#define Rep(i,a,b) for(int i = a; i < b; i++)
#define rep(i,b) Rep(i,0,b)
#define rrep(i,a) for(int i = a; i >= 0; i--)
#define allof(a) (a).begin(), (a).end()

typedef long long ll;
const int inf = 1e9 + 7;
const ll infll = 1ll << 60ll;
const ll mod = 1e9 + 7;
// 0~3までは右左下上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, -1, 0, 1, 1, -1, -1, 1 };

/*
// 最大公約数
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}

// 最小公倍数
ll lcm(ll a, ll b) {
	return a * b / gcd(a, b);
}*/


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n;
	cin >> n;
	vector<int> b(n);
	rep(i, n) cin >> b[i];

	vector<int> ans;

	while(n > 0) {
		for(int j = n - 1; j >= 0; j--){
			if (b[j] > (j + 1)) {
				cout << -1 << endl;
				return 0;
			}
			if (b[j] == (j + 1)) {
				ans.push_back(b[j]); // b[j]を抜き出す
				n--;
				Rep(k, j, n) { // 前に詰める
					b[k] = b[k + 1];
				}
				break;
			}
		}
	}

	for (int i = ans.size() - 1; i >= 0; i--) {
		cout << ans[i] << endl;
	}

	return 0;
}