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 <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <cstdio>
#include <utility>
#include <string>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <numeric>

using namespace std;

typedef uint64_t u64;
typedef int64_t s64;
typedef uint32_t u32;
typedef int32_t s32;
typedef vector<s32> vs32;
typedef vector<u32> vu32;
typedef vector<s64> vs64;
typedef vector<u64> vu64;

const double PI=3.14159265358979323846;

#define MAX(x, y) ((x) < (y) ? (y) : (x))
#define MIN(x, y) ((x) > (y) ? (y) : (x))

#define rep(i, N) for(int i = 0; i < N; ++i)

#define CEIL(x, y) (((x) + (y) - 1) / (y))
#define MOD 1000000007ULL

#define IN(l, r, x) ((l) <= (x) && (x) < (r))


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

    int n;
    cin >> n;
    vs32 b(n);
    rep (i, n)
    {
        cin >> b[i];
        if (b[i] > i + 1)
        {
            cout << "-1\n";
            return 0;
        }
    }

    vs32 ans(n);
    for (int i = n - 1; i >= 0; --i)
    {
        int pos = b.size() - 1;
        while (b[pos] != pos + 1) --pos;
        ans[i] = pos + 1;
        b.erase(b.begin() + pos);
    }

    rep (i, n)
    {
        cout << ans[i] << "\n";
    }
    return 0;
}