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;
#define For(i,n,k) for(ll i=(n);i<(k);i++)
ll n,num=0;
int graph[100][100]={0};
void Main(){    
    cin>>n;
    if(n==3){
        graph[0][2]=1;
        graph[1][2]=1;
        num=2;
    }
    else if(n%2){
        For(i,0,n){
            if(i==0||i==n-2){
                graph[i][n-1]=1;
                graph[i][1]=1;
                graph[i][n-3]=1;
                num+=3;
            }
            else if(i==n-1){
                graph[i][0]=1;
                graph[i][n-2]=1;
                graph[i][n/2]=1;
                graph[i][n/2-1]=1;
                num+=4;
            }
            else if(i==n/2||i==n/2-1){
                graph[i][n/2-2]=1;
                graph[i][n/2+1]=1;
                graph[i][n-1]=1;        
                num+=3;        
            }
            else{
                graph[i][i+1]=1;
                graph[i][i-1]=1;
                graph[i][n-i]=1;
                graph[i][n-i-2]=1;           
                num+=4;     
            }
        }
        num/=2;
    }
    else{
        For(i,0,n){
            if(i==0||i==n-1){
                graph[i][n/2]=1;
                graph[i][n/2-1]=1;
                graph[i][n-2]=1;
                graph[i][1]=1;
            }
            else if(i==n/2||i==n/2-1){
                graph[i][0]=1;
                graph[i][n-1]=1;
                graph[i][n/2-2]=1;     
                graph[i][n/2+1]=1;          
            }
            else{
                graph[i][i+1]=1;
                graph[i][i-1]=1;
                graph[i][n-i]=1;
                graph[i][n-i-2]=1;               
            }
        }
        if(n==4) num=4;
        else num=n*2;
    }
    cout<<num<<endl;
    For(i,0,n){
        for(ll j=i+1;j<n;j++){
            if(graph[i][j]) cout<<i+1<<" "<<j+1<<endl;
        }
    }
}
int main(){
    Main();
    return 0;
}