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 ll long long int
#define show(x) cout<<x<<'\n'
#define go(i,a,n) for(ll i=a;i<n;i++)
#define pb push_back
#define mod 1000000007
int main() {
    ll n;
    cin>>n;
    int a[n],b[n];
    map<int,int>m;
    ll flag;
    for(ll i=0;i<n;i++)
    {
        cin>>a[i];
        m[a[i]]++;
    }
    for(ll i=0;i<n;i+=1)
    {
        flag=0;
        for(ll j=1;j<=i+1;j++)
        {
            if(m[j]>0)
            {
                b[i]=j;
                m[j]--;
                flag=1;
                break;
            }
        }
        if(flag==0)
        {
            show("-1");
            exit(0);
        }
    }
    for(ll i=0;i<n;i++)
    {
        show(b[i]);
    }
}