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 Lua (Lua 5.3.5) (judge verdict: Wrong Answer)

local mfl, mce = math.floor, math.ceil
local n = io.read("*n")
local a = {}
for i = 1, n do
  a[i] = io.read("*n")
end
local xor = 0
for i = 3, n do
  xor = xor ~ a[i]
end
local sum = a[1] + a[2]
if sum < xor then
  print(-1)
  os.exit()
end
local b_xor, b_sum = {}, {}
while 0 < sum do
  table.insert(b_sum, sum % 2)
  table.insert(b_xor, xor % 2)
  sum = mfl(sum / 2)
  xor = mfl(xor / 2)
end
local dig = #b_sum
local moveup = false
local state = {}
for i = 1, dig do state[i] = 0 end
-- state of bit of a[1] and a[2]
-- 1: [0, 0], 2: [1, 1], 3: [0, 1] or [1, 0]
for idig = dig, 1, -1 do
  if b_sum[idig] == 1 then
    if b_xor[idig] == 1 then
      if moveup then
        print(-1)
        os.exit()
      else
        state[idig] = 3
      end
    else
      if moveup then
        state[idig] = 2
      else
        state[idig] = 1
      end
      moveup = true
    end
  else-- sum = 0
    if b_xor[idig] == 1 then
      if moveup then
        state[idig] = 3
      else
        print(-1)
        os.exit()
      end
    else
      if moveup then
        state[idig] = 2
      else
        state[idig] = 1
      end
      moveup = false
    end
  end
end
local alim = {}
do
  local tmp = a[1]
  for i = 1, dig do
    alim[i] = tmp % 2
    tmp = mfl(tmp / 2)
  end
end
local tbl = {}
for i = 1, dig do tbl[i] = 0 end

local function dfs(idig, limit)
  if idig == 0 then return true end
  if limit then
    if alim[idig] == 0 then
      if state[idig] == 2 then
        return false
      else
        tbl[idig] = 0
        return dfs(idig - 1, true)
      end
    else
      if state[idig] == 2 then
        tbl[idig] = 1
        return dfs(idig - 1, true)
      elseif state[idig] == 1 then
        tbl[idig] = 0
        return dfs(idig - 1, false)
      else
        tbl[idig] = 1
        local ret = dfs(idig - 1, true)
        if ret then return true
        else
          tbl[idig] = 0
          return dfs(idig - 1, false)
        end
      end
    end
  else
    if state[idig] == 1 then
      tbl[idig] = 0
      return dfs(idig - 1, false)
    else
      tbl[idig] = 1
      return dfs(idig - 1, false)
    end
  end
end

local ret = dfs(dig, true)
if not ret then
  print(-1)
  os.exit()
end
local ans = 0
local mul = 1
for i = 1, dig do
  ans = ans + mul * tbl[i]
  mul = mul * 2
end
if ans == 0 then
  print(-1)
else
  print(a[1] - ans)
end