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;

#define INF 1000007
#define MOD 1000000007
#define maxn 100
#define int long long int
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define rep(i, n) FOR(i, 0, n)

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

int lcm(int a, int b) { return a / gcd(a, b) * b; }

int mpow(int N, int P, int M)  //繰り返し自乗法
{
    if (P == 0)
        return 1;
    if (P % 2 == 0) {
        int t = mpow(N, P / 2, M);
        return t * t % M;
    }
    return N * mpow(N, P - 1, M) % M;
}

vector<bool> prime(maxn + 1, true); //素数テーブル
void prime_table(void) {
    prime[0] = false;
    prime[1] = false;
    for (int i = 2; i * i <= maxn; i++) {
        if (!prime[i]) {
            continue;
        }
        for (int j = i + i; j <= maxn; j += j + i) {
            prime[j] = false;
        }
    }
}

signed main() {
    int N;
    cin >> N;
    vector<int> b;
    priority_queue<int, vector<int>, greater<int>> que;
    vector<int> ans;
    rep(i, N) {
        int a;
        cin >> a;
        b.push_back(a);
    }
    for (int i = N - 1; i >= 0; i--) {
        que.push(b[i]);
        if (b[i] == 1) {
            while (!que.empty()) {
                ans.push_back(que.top());
                que.pop();
            }
        }
    }
    if (true) {
        cout << "-1" << endl;
    } else {
        rep(i, ans.size()) {
            cout << ans[i] << endl;
        }

    }
}