p2356_鸽笼原理！cool！

Find a multiple

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1895		Accepted: 813		Special Judge

Description

The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000. This numbers are not necessarily different (so it may happen that two or more of them will be equal). Your task is to choose a few of given numbers ( 1 <= few <= N ) so that the sum of chosen numbers is multiple for N (i.e. N * k = (sum of chosen numbers) for some natural number k).

Input

The first line of the input contains the single number N. Each of next N lines contains one number from the given set.

Output

In case your program decides that the target set of numbers can not be found it should print to the output the single number 0. Otherwise it should print the number of the chosen numbers in the first line followed by the chosen numbers themselves (on a separate line each) in arbitrary order.

If there are more than one set of numbers with required properties you should print to the output only one (preferably your favorite) of them.

Sample Input

5



1



2



3



4



1

Sample Output

2



2



3

Source

Ural Collegiate Programming Contest 1999

非常酷的一道题；

n个数中找出一些数可以整除n，就是找都某几个数的和对n取余得0；

设s[k]是前k的和，那么在所有n个s中，必定有一个为0 或者存在两个相等。

这就应用了传说中的鸽笼原理。因为对n取余的结果是0---n-1放在n个s中，至少有两个相等，那么相等的位置之间的和对n取余的结果一定是0。

后记：很酷吧！不知道是谁想出来的。。

前几天刚看过鸽笼原理，觉得确实很好玩，但是碰到这道题后就是没有想到，有些信息确实很有用的，首先n很大对不，n^2肯定不行；第二为什么非要等于n呢，您说呢～，数学这东东很有意思！