## Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs *N* (1 ≤ *N* ≤ 20,000) planks of wood, each having some integer length *Li* (1 ≤ *Li* ≤ 50,000) units. He then purchases a single long board just long enough to saw into the *N* planks (i.e., whose length is the sum of the lengths *Li*). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; you should ignore it, too.

FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the *N*-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the *N* planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.

## Input

Line 1: One integer *N*, the number of planks Lines 2..*N*+1: Each line contains a single integer describing the length of a needed plank

## Output

Line 1: One integer: the minimum amount of money he must spend to make *N*-1 cuts

## Sample Input

1 | 3 |

## Sample Output

1 | 34 |

## Hint

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8. The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

## Answer

1 | `#include <cstdio>` |

## Analysis

- 哈夫曼编码，每次取最短的两个木板合并，其合成（剪开）的开销最小。
- 在 while 循环中，每一轮选取最短的两个进行合并，然后合并结果放入剩下的木板中重新排序，在下一轮中依旧选择剩下的最小的两块木板。
- 如果每一轮都快排重新排序，超时，这里才有常数时间的插入方法（因为数组已经是个有序数组）。相当于在有序数组中插入一个数，时间复杂度为 O(N)。
- 注意范围，答案是 long long 类型。
- 采用优先队列的方法，每次取出最小的两个，并插入它们长度之和，直至剩下最后一块木板。