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;

typedef long long ll;
const ll MOD = 1e9 + 7;
const ll INF = 1e18;

int main() {
    ll n, m, i, j, k;
    cin>>n;
    vector<ll> b(n);
    for(i=0;i<n;i++){
        cin >> b[i];
    }
    ll bs=n;
    vector<ll> ret;
    while(bs>0){
        ll now = -1;
        for(i=0;i<bs;i++){
            if(i+1==b[i])now = i;
        }
        if(now>=0){
            ret.push_back(now+1);
            b.erase(b.begin()+now);
            bs--;
        }else{
            cout << -1 << endl;
            return 0;
        }
    }
    reverse(ret.begin(),ret.end());
    for(ll x:ret){
        cout << x << endl;
    }
    return 0;
}