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;
#define int long long
#define pb push_back
#define fi first
#define se second
#define FOR(i,n) for(int i = 0;i<n;i++)
#define rep(i,s,n) for(int i = s;i<n;i++)
#define rrep(i,s,n) for(int i = (n)-1;i>=(s);i--)
#define all(v) (v).begin(),(v).end()
#define chmin(a,b) a=min((a),(b))
#define chmax(a,b) a=max((a),(b))
#define endl '\n'
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0)
typedef long long ll;
typedef pair<int,int>pint;
typedef vector<int>vint;
typedef vector<pint>vpint;
const ll MOD=998244353,INF=1ll<<60;
 
int N;
 
signed main() {
    IOS();
    cin>>N;
    vint a;
    rep(i,0,N){
        int t;
        cin>>t;
        t--;
        if(t>i){
            cout<<-1<<endl;
            return 0;
        }
        a.pb(t);
    }
    
    vint ans;
    rep(i,0,N){
        int sz=(int)a.size();
        int ma=-1;
        rep(j,0,sz){
            if(a[j]==j)chmax(ma,j);
        }
        ans.pb(ma);
        a.erase(a.begin()+ma);
    }
    
    
    rrep(i,0,(int)ans.size()){
        cout<<ans[i]+1<<endl;
    }
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    
    return 0;
}