This is a problem from SPOJ. I am getting W/A on submission.The sample test case is passing. I am looking for some test-cases where my code will fail. If anyone could help me with that, It will be a huge help.

Ada the Ladybug is playing Game of Divisors against her friend Velvet Mite Vinit. The game has following rules. There is a pile of N stones between them. The player who’s on move can pick at least 1 an at most σ(N) stones (where σ(N) stands for number of divisors of N). Obviously, N changes after each move. The one who won’t get any stones (N == 0) loses.

As Ada the Ladybug is a lady, so she moves first. Can you decide who will be the winner? Assume that both players play optimally.

Input The first line of input will contain 1 ≤ T ≤ 10^5, the number of test-cases.

The next T lines will contain 1 ≤ N ≤ 2*107, the number of stones which are initially in pile.

Output Output the name of winner, so either “Ada” or “Vinit”.

Example Input
8
1
3
5
6
11
1000001
1000000
29
Example Output:
Ada
Vinit
Ada
Ada
Vinit
Vinit
Ada
Ada

CODE :

import java.io.*;  public class Main {     public static int max_size = 2 * (int)Math.pow(10,7) + 1;     public static boolean[] dp = new boolean[max_size];     public static int[] lastPrimeDivisor = new int[max_size];     public static int[] numOfDivisors = new int[max_size];     public static void main(String[] args) throws IOException     {         BufferedReader br = new BufferedReader(new InputStreamReader(System.in));         preprocess();         int t = Integer.parseInt(br.readLine());         while(t > 0)         {             int n = Integer.parseInt(br.readLine());             if(dp[n] == true)                 System.out.println("Ada");             else                 System.out.println("Vinit");             t--;         }     }     public static void markLastPrimeDivisor()     {         for(int i = 0 ; i < max_size ; i++)         {             lastPrimeDivisor[i] = 1;         }         for(int i = 2 ; i < max_size ; i += 2)         {             lastPrimeDivisor[i] = 2;         }         int o = (int)Math.sqrt(max_size) + 1;         for(int i = 3 ; i < max_size; i++)         {             if(lastPrimeDivisor[i] != 1)             {                 continue;             }             lastPrimeDivisor[i] = i;             if(i <= o)             {                 for(int j = i * i ; j < max_size ; j += 2 * i)                 {                     lastPrimeDivisor[j] = i;                 }             }         }         /*for(int i = 1 ; i < max_size ; i++)             System.out.println("last prime of " + i + " is " + lastPrimeDivisor[i]);*/     }      public static void countDivisors(int num)     {         int original = num;         int result = 1;         int currDivisorCount = 1;         int currDivisor = lastPrimeDivisor[num];         int nextDivisor;         while(currDivisor != 1)         {             num = num / currDivisor;             nextDivisor = lastPrimeDivisor[num];             if(nextDivisor == currDivisor)             {                 currDivisorCount++;             }             else             {                 result = result * (currDivisorCount + 1);                 currDivisorCount = 1;                 currDivisor = nextDivisor;             }         }         if(num != 1)         {             result = result * (currDivisorCount + 1);         }         //System.out.println("result for num : " + original + ", " + result);         numOfDivisors[original] = result;     }      public static void countAllDivisors()     {         markLastPrimeDivisor();         for(int i = 2 ; i < max_size ; i++)         {             countDivisors(i);             //System.out.println("num of divisors of " + i + " = " + numOfDivisors[i]);         }     }       public static void preprocess()     {         countAllDivisors();         dp[0] = dp[1] = dp[2] = true;         for(int i = 3 ; i < max_size ; i++)         {             int flag = 0;             int limit = numOfDivisors[i];             for(int j = 1 ; j <= limit; j++)             {                 //If for any i - j, we get false,for playing optimally                 //the current opponent will choose to take j stones out of the                 //pile as for i - j stones, the other player is not winning.                 if(dp[i - j] == false)                 {                     dp[i] = true;                     flag  = 1;                     break;                 }             }             if(flag == 0)                 dp[i] = false;         }      } }