def lscan; gets.split.map(&:to_i); end
def bye(msg); puts msg.to_s; exit; end
def 🎍(a,b,c)
  a != c && ((a < b && b > c) || (a > b && b < c))
end

def solve1(a,b,c) # bmax
  cost = 0
  if b <= c
    cost += c - b + 1
    c = b - 1
  end
  if b <= a
    cost += a - b + 1
    a = b - 1
  end
  if a == c
    cost += 1
    a -= 1
  end
  return a > 0 && b > 0 && c > 0 && cost
end

def solve2(a,b,c) # bmin
  cost = 0
  if b >= c
    cost += b - c + 1
    b = c - 1
  end
  if b >= a
    cost += b - a + 1
    b = a - 1
  end
  if a == c
    if a-b > 1
      a -= 1
      cost += 1 
    else
      a -= 1
      x = a < b ? 2 : 1
      b -= x
      cost += 1 + x
    end
  end
  #p [a,b,c, cost]
  return a > 0 && b > 0 && c > 0 && cost
end

def query(a,b,c)
  return 0 if 🎍(a,b,c)
  a,c = c,a if a > c
  s1 = solve1(a,b,c)
  s2 = solve2(a,b,c)
  return s1 && s2 ? [s1, s2].min : s1 || s2 ? s1 || s2 : -1
end

gets.to_i.times do
  p query(*lscan)
end