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: Wrong Answer)

#include<bits/stdc++.h>

using namespace std;
using ll = long long;

#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define debug(x) cout << #x << ": " << x << endl;

int main() {
    int N;
    cin >> N;
    vector<int> b(N);
    map<int, int> cnt;
    for (int i = 0; i < N; ++i) {
        cin >> b[i];
        b[i]--;
        cnt[b[i]]++;
        if (b[i] >= N or b[i] > i) {
            cout << -1 << endl;
            return 0;
        }
    }

    if (b[0] != 0) {
        cout << -1 << endl;
        return 0;
    }

    for (int i = 0; i < N; i++) {
        if (cnt[i] > N - i) {
            cout << -1 << endl;
            return 0;
        }
    }

    vector<int> ans(N);
    vector<bool> flg(N, false);
    ans[0] = 1;
    flg[0] = true;
    int p = N - 1, idx = N - 1;
    while (p > 0) {
        if (flg[idx]) {
            idx--;
            continue;
        }
        if (b[idx] >= N - p) {
            int loop = b[idx] - N + p + 1;
            for (int k = 0; k < loop; k++) {
                for (int j = idx - 1; j > 0; j--) {
                    if (flg[j])continue;
                    if (b[j] <= N - p) {
                        ans[N - p] = b[j] + 1;
                        flg[j] = true;
                        p--;
                        break;
                    }
                }
            }
        }
        ans[N - p] = b[idx] + 1;
        flg[idx] = true;
        idx--;
        p--;
    }

    for (int i = 0; i < N; i++)cout << ans[i] << endl;

    return 0;
}