ZSoft Corp. is a software company in GaoKe Hall. And the workers in the hall are very hard-working. But the elevator in that hall always drives them crazy. Why? Because there is only one elevator in GaoKe Hall, while there are hundreds of companies in it. Every morning, people must waste a lot of time waiting for the elevator.

Hal, a smart guy in ZSoft, wants to change this situation. He wants to find a way to make the elevator work more effectively. But it's not an easy job.

There are 31 floors in GaoKe Hall. It takes 4 seconds for the elevator to raise one floor. It means:

It costs (31 - 1)×4 = 120 seconds if the elevator goes from the 1-st floor to the 31-st floor without stop. And the elevator stops 10 second once. So, if the elevator stops at each floor, it will cost 30×4 + 29×10 = 410 seconds (It is not necessary to calculate the stopping time at 31st floor). In another way, it takes 20 seconds for the workers to go up or down one floor. It takes 30×20 = 600 seconds for them to walk from the 1-st floor to the 31-st floor. Obviously, it is not a good idea. So some people choose to use the elevator to get a floor which is the nearest to their office.

After thinking over for a long time, Hal finally found a way to improve this situation. He told the elevator man his idea: First, the elevator man asks the people which floors they want to go. He will then design a stopping plan which minimize the time the last person need to arrive the floor where his office locates. For example, if the elevator is required to stop at the 4-th, 5-th and 10-th floor, the stopping plan would be: the elevator stops at 4-th and 10-th floor. Because the elevator will arrive 4^th floor at 3×4 = 12 second, then it will stop 10 seconds, then it will arrive 10^th floor at 3×4 + 10 + 6×4 = 46 second. People who want to go 4-th floor will reach their office at 12 second, people who want to go to 5-th floor will reach at 12 + 20 = 32 second and people who want to go to 10-th floor will reach at 46 second. Therefore it takes 46 seconds for the last person to reach his office. It is a good deal for all people.

Now, you are supposed to write a program to help the elevator man to design the stopping plan, which minimize the time the last person needs to arrive at his floor.

Input 

The input consists of several testcases. Each testcase is in a single line as the following:

It means, there are totally n floors at which the elevator need to stop, and n = 0 means no testcases any more. f₁ f₂ ... f_n are the floors at which the elevator is to be stopped (n30, 2f₁ < f₂ < ... < f_n31). Every number is separated by a single space.

Output 

For each testcase, output the time the last reading person needs in the first line and the stopping floors in the second line. Please note that there is a summary of the floors at the head of the second line. There may be several solutions, any appropriate one is accepted. No extra spaces are allowed.

Sample Input 

3 4 5 10
1 2
0

Sample Output 

46
2 4 10
4
1 2

二分答案，當然也可以選擇 DP，dp[i][j] 表示電梯當前停留在 i 時，把 j 個人送到指定樓層的最少時間。

當指定一個時間後，檢查在限定的時間內能不能將所有人送到指定樓層的策略。

貪婪策略：電梯盡可能在越高的地方停留，即加上要往下爬的那個人不超過指定時間。

因此可以得到當電梯停留在 j 層時，而有一個人目標 i 樓層，此時電梯已經停留 num 次。

則必須符合 (j'-1)*4 + num*10 + (j'-i)*20 <= limit_time ==> 得到 j' 下次停留地方。

而同時相對應的，這個時候放人並不是只能往下爬，往上爬也是等價的。

因此可以忽略中間一段人的需求，得到下次考慮的人的目標樓層。

(j'-1)*4 + num*10 + (i'-j')*10 <= limit_time ==> i' (含)以下的不用考慮。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std;
int f[105];
vector<int> stopFloor;
int check(int limit, int mx) {
	int lfloor = limit/20+2;// floor of last person be stopped.
	int sfloor = 1;// now stopping floor.
	int scnt = 0;// stopped times.
	vector<int> buf;
	while(lfloor <= mx) {
		while(lfloor <= mx && f[lfloor] == 0)
			lfloor++;
		if(scnt*10+(lfloor-1)*4 > limit)
			return 0;
		int nfloor = (limit-10*scnt+20*lfloor+4)/24;//next stopping floor.
		lfloor = (limit-10*scnt+16*nfloor+4)/20+1;
		sfloor = nfloor;
		//if(nfloor <= mx)
		buf.push_back(nfloor);
		scnt++;
	}
	stopFloor = buf;
	return 1;
}
int main() {   
	//freopen("in.txt","r+t",stdin);
    //freopen("out.txt","w+t",stdout);
	int n;
	int i, j, k, x;
	while(scanf("%d", &n) == 1 && n) {
		memset(f, 0, sizeof(f));
		int mx = 0;
		for(i = 0; i < n; i++) {
			scanf("%d", &x), f[x] = 1;
			mx = max(mx, x); 
		}
		int l = 0, r = mx*(14), mid;
		int ret = 0;
		while(l <= r) {
			mid = (l+r)/2;
			if(check(mid, mx))
				ret = mid, r = mid-1;
			else
				l = mid+1;
		}
		printf("%d\n", ret);
		printf("%d", stopFloor.size());
		for(i = 0; i < stopFloor.size(); i++)
			printf(" %d", stopFloor[i]);
		puts("");
	}
	return 0;
}