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: Wrong Answer)

import java.util.*;

public class Main {
	static final int LEN = 41;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int N = sc.nextInt();
		long[] A = new long[N];
		for(int i=0; i<N; i++) {
			A[i] = sc.nextLong();
		}
		
		long goal = 0;
		for(int i=2; i<N; i++)
			goal ^= A[i];
		
		System.out.println(solve(A[0], A[1], goal));
		sc.close();
	}
	
	static long solve(long a0, long a1, long goal) {
		if((a0^a1)==goal)
			return 0;
		
		long sum = a0+a1;
		
		int[] bits = new int[LEN];
		boolean carry = false;

		for(int i=LEN-1; i>=0; i--) {
			if(carry) {
				if(bit(sum, i)==0) {
					if(bit(goal, i)==0) {
						bits[i] = 2;
						carry = false;
					} else {
						bits[i] = 1;
					}
				} else {
					if(bit(goal, i)==0) {
						bits[i] = 2;
					} else {
						return -1;
					}
				}
			} else {
				if(bit(sum, i)==0) {
					if(bit(goal, i)==0) {
						bits[i] = 0;
					} else {
						return -1;
					}
				} else {
					if(bit(goal, i)==0) {
						bits[i] = 0;
						carry = true;
					} else {
						bits[i] = 1;
					}
				}
			}
		}

		long ans = Long.MAX_VALUE;
		for(int i=LEN-1; i>=0; i--) {
			if(bit(a0, i)==1) {
				boolean ok = true;
				long after = 0;
				for(int j=LEN-1; j>=0; j--) {
					if(j>i) {
						if((bit(a0, j)==0 && bits[j]==2)||(bit(a0, j)==1 && bits[j]==0)) {
							ok = false;
							break;
						}
						after |= ((long)(bit(a0, j)))<<j;
					} else if(j==i) {
						if(bits[j]==2) {
							ok = false;
							break;
						}
					} else {
						if(bits[j]>0) {
							after |= 1l<<j;
						}
					}
				}
				if(ok) {
					ans = Math.min(ans, a0-after);
				}
			}
		}
		
		return ans < Long.MAX_VALUE ? (ans==a0 ? -1 : ans) : -1;
	}
	
	static int bit(long a, int b) {
		return (int)((a>>b)&1);
	}
}