Problem:
Score : 600 points

Problem Statement

There are N piles of stones. The i-th pile has A_i stones.

Aoki and Takahashi are about to use them to play the following game:

Starting with Aoki, the two players alternately do the following operation:

Operation: Choose one pile of stones, and remove one or more stones from it.

When a player is unable to do the operation, he loses, and the other player wins.

When the two players play optimally, there are two possibilities in this game: the player who moves first always wins, or the player who moves second always wins, only depending on the initial number of stones in each pile.

In such a situation, Takahashi, the second player to act, is trying to guarantee his win by moving at least zero and at most (A_1 - 1) stones from the 1-st pile to the 2-nd pile before the game begins.

If this is possible, print the minimum number of stones to move to guarantee his victory; otherwise, print -1 instead.

Constraints

2 \leq N \leq 300

1 \leq A_i \leq 10^{12}

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

Print the minimum number of stones to move to guarantee Takahashi's win; otherwise, print -1 instead.

Sample Input 1

2
5 3

Sample Output 1

1

Without moving stones, if Aoki first removes 2 stones from the 1-st pile, Takahashi cannot win in any way.

If Takahashi moves 1 stone from the 1-st pile to the 2-nd before the game begins so that both piles have 4 stones, Takahashi can always win by properly choosing his actions.

Sample Input 2

2
3 5

Sample Output 2

-1

It is not allowed to move stones from the 2-nd pile to the 1-st.

Sample Input 3

3
1 1 2

Sample Output 3

-1

It is not allowed to move all stones from the 1-st pile.

Sample Input 4

8
10 9 8 7 6 5 4 3

Sample Output 4

3

Sample Input 5

3
4294967297 8589934593 12884901890

Sample Output 5

1

Watch out for overflows.

Solution in Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
      	int n = sc.nextInt();
      	long[] num = new long[n];
      	for(int i=0;i<n;i++){
          num[i] = sc.nextLong();
        }
      	
  long max = 0;
  boolean Noans = true;
  for(int i=0;i<num.length;i++){
    if( num[i] > max ){
      max = num[i];
    }
  }
  String s = Long.toBinaryString(max);
  boolean[] d = new boolean[ s.length() ];
  for(int i=0;i<d.length;i++){
    d[i] = true;
  }
  for(int i=2;i<num.length;i++){
    s = Long.toBinaryString( num[i] );
    for(int j=0;j<s.length();j++){
      int t = d.length - s.length() + j;
      if( ( d[t] && s.charAt(j) == '0' ) || ( !d[t] && s.charAt(j) == '1' ) ){
        d[t] = true;
      }else{
        d[t] = false;
      }
      //System.out.print( d[t] );
    }
    //System.out.println("ここまで"+i);
  }
  
  //System.out.println(a1+":"+a2+":"+d[0]+d[1]);
  
  for(int i=0;i<num[0];i++){
    String s1 = Long.toBinaryString( num[0] );
    String s2 = Long.toBinaryString( num[1] );
    boolean [] tmp = new boolean[d.length];
    for(int j=0;j<tmp.length;j++){
      tmp[j] = d[j];
    }
    
    for(int j=0;j<s1.length();j++){
      int t = d.length - s1.length() + j;
      if( ( tmp[t] && s1.charAt(j) == '0' ) || ( !tmp[t] && s1.charAt(j) == '1' ) ){
        tmp[t] = true;
      }else{
        tmp[t] = false;
      }
    }
    for(int j=0;j<s2.length();j++){
      int t = d.length - s2.length() + j;
      if( ( tmp[t] && s2.charAt(j) == '0' ) || ( !tmp[t] && s2.charAt(j) == '1' ) ){
        tmp[t] = true;
      }else{
        tmp[t] = false;
      }
    }
    boolean ans = true;
    for(int j=0;j<d.length;j++){
      if( !tmp[j] ){
        ans = false;
      }
    }
    if( ans ){
      System.out.println(i);
      Noans = false;
      break;
    }
    num[0]--;
    num[1]++;
    
  }
  if(Noans){
      System.out.println(-1);
    }

	}
}