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 <bits/stdc++.h>
using namespace std;
using lint = int64_t;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define all(v)    begin(v), end(v)
#define in        []{lint n; cin>>n; return n;}()

int main(){
	int N = in;
	vector<int> A(N);
    rep( i, N ) A[i] = in-1;

    stack<int> ans;
    while( !A.empty() ) {
        for( int i=A.size()-1; i>=0; i-- ) {
            if( A[i] == i ) {
                A.erase( A.begin() +i );
                ans.push(i);
                goto E;
            }
        }
        cout << "-1\n";
        return 0;
        E:;
    }

    while( !ans.empty() ) {
        cout << ans.top()+1 << "\n";
        ans.pop();
    }
}