HDU-1394 Minimum Inversion Number (逆序数，线段树或树状数组)

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. 



For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following: 



a1, a2, ..., an-1, an (where m = 0 - the initial seqence) 

a2, a3, ..., an, a1 (where m = 1) 

a3, a4, ..., an, a1, a2 (where m = 2) 

... 

an, a1, a2, ..., an-1 (where m = n-1) 



You are asked to write a program to find the minimum inversion number out of the above sequences. 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1. 

Output

For each case, output the minimum inversion number on a single line. 

Sample Input

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

逆序数，就是你比人家小，结果还在人家后边。。。可以用线段树或树状数组求，每一次出现一个数，设为a，就求一次a到n之间的和，也就是比a大却先比a出现的数字个数，然后把a的位置标记为一，其余几次的逆序数可以直接推导出来，设第一次移动的数为a，把它移动到后面理论上会减少a个逆序数(注意从0开始)，但是还会有n-1-a个数大于a，所以sum+=n-1-2*a;

线段树:

zkw线段树: