You want to processe a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. Then how many times it need. 

For example, 1 2 3 5 4, we only need one operation : swap 5 and 4. 

InputThe input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 1000); the next line contains a permutation of the n integers from 1 to n.OutputFor each case, output the minimum times need to sort it in ascending order on a single line.

Sample Input

31 2 34 4 3 2 1

Sample Output

06 大概意思：给有序数组里相邻的数字进行排序，算出排序时需要进行的最小步骤数。

1 #include 
     
       2 #define ll long long 3 using namespace std; 4 int a[1010]; 5 int main(){ 6     int n; 7     while(scanf("%d",&n)!=EOF){ 8         for(int i = 0; i < n; i ++){ 9             //scanf("%d",a[i]);10             cin>>a[i];11         }12         ll ans = 0;13         for(int i =1; i < n; i ++){
     //13到17行为冒泡排序法14             for(int j = 0; j 
      
       a[j+1]){16                     ans++;17                     swap(a[j],a[j+1]);18                 }19             }20         }21         cout << ans << endl;22         //printf("")23     }24     return 0;25 }