a = read_line.split.map(&.to_i)
b2 = 0
b3 = 0
3.times do |i|
  b2 |= 1 << i if a[i] % 2 == 0
  b3 |= 1 << i if a[i] % 3 == 0
end
puts b3.popcount > 0 && (b2 & ~b3).popcount > 0 ? "Yes" : "No"