Problem:
Score : 1000 points

Problem Statement

Snuke participated in a magic show.

A magician prepared N identical-looking boxes.
He put a treasure in one of the boxes, closed the boxes, shuffled them, and numbered them 1 through N.

Since the boxes are shuffled, now Snuke has no idea which box contains the treasure.
Snuke wins the game if he opens a box containing the treasure.
You may think that if Snuke opens all boxes at once, he can always win the game.
However, there are some tricks:

Snuke must open the boxes one by one. After he opens a box and checks the content of the box, he must close the box before opening the next box.

He is only allowed to open Box i at most a_i times.

The magician may secretly move the treasure from a closed box to another closed box, using some magic trick.
For simplicity, assume that the magician never moves the treasure while Snuke is opening some box.
However, he can move it at any other time (before Snuke opens the first box, or between he closes some box and opens the next box).

The magician can perform the magic trick at most K times.

Can Snuke always win the game, regardless of the initial position of the treasure and the movements of the magician?

Constraints

2 \leq N \leq 50

1 \leq K \leq 50

1 \leq a_i \leq 100

Input

Input is given from Standard Input in the following format:

N K
a_1 a_2 \cdots a_N

Output

If the answer is no, print a single -1.

Otherwise, print one possible move of Snuke in the following format:

Q
x_1 x_2 \cdots x_Q

It means that he opens boxes x_1, x_2, \cdots, x_Q in this order.

In case there are multiple possible solutions, you can output any.

Sample Input 1

2 1
5 5

Sample Output 1

7
1 1 2 1 2 2 1

If Snuke opens the boxes 7 times in the order 1 \rightarrow 1 \rightarrow 2 \rightarrow 1 \rightarrow 2 \rightarrow 2 \rightarrow 1, he can always find the treasure regardless of its initial position and the movements of the magician.

Sample Input 2

3 50
5 10 15

Sample Output 2

-1

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using static System.Console;
using static System.Math;
 
public class Hello{
    static public int mod = 1000000007;
    static public long k = 0;
    public static void Main(){
        // 方針
        int[] num = inta();
        int[] op = inta();
        int min = 99999;
        for(int i=0;i<num[0];i++){
            min = Min(op[i],min);
        }
        int[] minlist = new int[num[0]];
        int mincount = 0;
        for(int i=0;i<num[0];i++){
            if(op[i] == min){
                minlist[i]++;
                mincount++;
            }
        }
        int canfind = 0;
        if(mincount == 1){
            if(num[1] <= 2*min-1){
                canfind = 1;
            }
        }else if(mincount >= 2){
            if(num[1] <= 2*min-2){
                canfind = 2;
            }
        }
        
        if(canfind == 0){
            WriteLine(-1);
        }else{
            WriteLine(num[0]*num[1]);
            for(int i=0;i<min;i++){
                for(int j=1;j<=num[0];j++){
                    if(minlist[j-1] == 0){
                        Write($"{j} ");
                    }
                }
                for(int j=1;j<=num[0];j++){
                    if(minlist[j-1] == 1){
                        Write($"{j} ");
                    }
                }
            }
            WriteLine();
        }
        
        
        

        

        
    

        
        
        
        
    }
    static int ncr(int n,int r){r = Min(r,n-r);long nn = 1;for(int i=n-r+1;i<=n;i++){nn = nn*i%mod;}long rr = 1;for(int i=1;i<=r;i++){rr = rr*i%mod;}rr = square((int)rr,mod-2);nn = nn * rr %mod;return (int)nn;}
    // a^b mod
    static int square(int a,int b){string binary = Convert.ToString(b,2);int bileng = binary.Length;long a_power = a;long value = 1;for(int i=bileng-1;i>=0;i--){if(binary[i] == '1'){value = value*a_power%mod;}a_power = a_power*a_power%mod;}return (int)value;}
    //static int square2(long a,long b,long m){var list = new List<int>();int temp = 1;long n2 = a;list.Add(1);int count = 1;while(temp < b){list.Add((int)n2);temp*=2;count++;n2 = n2*n2%m;}long ret = 1;while(b > 0){if(temp <= b){b -= temp;ret = ret*list[count]%1000000007;}count--;temp/=2;}return(int)ret;}
    //各種読取
    static string rstr(){ return ReadLine(); }
    static int rint(){ return int.Parse(ReadLine()); }
    static long rlong(){ return long.Parse(ReadLine()); }
    static string[] stra(){ return ReadLine().Split(' '); }
    static int[] inta(){ string[] read_str_array = ReadLine().Split(' '); int[] return_int_array = new int[read_str_array.Length]; for(int countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_int_array[countup_i] = int.Parse(read_str_array[countup_i]); } return return_int_array; }
    static int[,] inta2(int num_array,int in_array){ int[,] int_array2 = new int[num_array,in_array]; for(int i=0;i<num_array;i++){ int[] temp_array = inta(); for(int j=0;j<in_array;j++){ int_array2[i,j] = temp_array[j]; } } return int_array2; }
    static long[] longa(){ string[] read_str_array = ReadLine().Split(' '); long[] return_long_array = new long[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_long_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_long_array; }
    static double[] doublea(){ string[] read_str_array = ReadLine().Split(' '); double[] return_double_array = new double[read_str_array.Length]; for(long countup_i=0;countup_i<read_str_array.Length;countup_i++){ return_double_array[countup_i] = long.Parse(read_str_array[countup_i]); } return return_double_array; }
    // -----------------------------
    static long divideup(long divided,long divid){ long temp_divide = divided/divid; if(divided % divid > 0){ temp_divide++; } return temp_divide; }
    static long GCD(long a,long b){ if(a < b){ long temp = a; a = b; b = temp; } if(a % b == 0){ return b; } else{ long temp = b; b = a%b; a = temp; return GCD(a,b); } }
    static long LCM(long a,long b){ return a * b / GCD(a,b); }
}