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;
typedef long long ll;
typedef long long int llint;
typedef pair<ll, ll> pa;
#define MM 1000000000
#define MOD MM+7
#define MAX 101000
#define MAP 110
#define initial_value -1
#define Pair pair<int,int>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
#define INF (1 << 29) //536870912
int dx[4] = {-1,0,1,0};
int dy[4] = {0,-1,0,1};
int N;
vector<int> b;
int main(){
    cin >> N;
    for(int i = 0; i < N; i++){
        int x; cin >> x;
        b.push_back(x);
        if(x > i+1){
            cout << -1 << endl;
            return 0;
        }
    }
    vector<int> a(0);
    while(b.size()){
        bool flg = false;
        for(int i = b.size()-1; i >= 0; i--){
            if(b[i] == i+1){
                a.push_back(b[i]);
                b.erase(b.begin() + i);
                flg = true;
                break;
            }
            if(!flg){
                cout << -1 << endl;
                return 0;
            }
        }
    }
    for(int i = a.size()-1; i >= 0; i--){
        cout << a[i] << endl;
    }
    return 0;
}