There is an old tradition of keeping 4 boxes of candies in the house in Cyberland. The numbers of candies are special if their arithmetic mean , their median and their range are all equal. By definition, for a set { x 1 , x 2 , x 3 , x 4 } ( x 1 ≤ x 2 ≤ x 3 ≤ x 4 ) arithmetic mean is , median is and range is x 4 - x 1 . The arithmetic mean and median are not necessary integer. It is well-known that if those three numbers are same, boxes will create a "debugging field" and codes in the field will have no bugs. For example, 1, 1, 3, 3 is the example of 4 numbers meeting the condition because their mean, median and range are all equal to 2 . Jeff has 4 special boxes of candies. However, something bad has happened! Some of the boxes could have been lost and now there are only n ( 0 ≤ n ≤ 4 ) boxes remaining. The i -th remaining box contains a i candies. Now Jeff wants to know: is there a possible way to find the number of candies of the 4 - n missing boxes, meeting the condition above (the mean, median and range are equal)? The first line of input contains an only integer n ( 0 ≤ n ≤ 4 ). The next n lines contain integers a i , denoting the number of candies in the i -th box ( 1 ≤ a i ≤ 500 ). In the first output line, print " YES " if a solution exists, or print " NO " if there is no solution. If a solution exists, you should output 4 - n more lines, each line containing an integer b , denoting the number of candies in a missing box. All your numbers b must satisfy inequality 1 ≤ b ≤ 10 6 . It is guaranteed that if there exists a positive integer solution, you can always find such b 's meeting the condition. If there are multiple answers, you are allowed to print any of them. Given numbers a i may follow in any order in the input, not necessary in non-decreasing. a i may have stood at any positions in the original set, not necessary on lowest n first positions . For the first sample, the numbers of candies in 4 boxes can be 1, 1, 3, 3 . The arithmetic mean, t