#include <iostream>
using namespace  std;
const int N=5e5+5;
typedef long long LL;
LL a[N],t[N];
int n;
LL merge(int l,int r)
{
    if(l>=r)return 0;
    int mid=l+r>>1;
    LL res=merge(l,mid)+merge(mid+1,r);

    int k=0;
    int i=l,j=mid+1;

    while(i<=mid&&j<=r)
    {
        if(a[i]>a[j]){res+=mid-i+1;t[k++]=a[j++];}
        else t[k++]=a[i++];
    }

    while(i<=mid) t[k++]=a[i++];
    while(j<=r) t[k++]=a[j++];

    for(int i=l,j=0;i<=r;i++,j++)a[i]=t[j];
    return res;
}


int main()
{

    while(cin>>n,n)  //正常读取为true   第2个n用于检测，若成功读取，读取的结果是否为0
    {
        for(int i=1;i<=n;i++)cin>>a[i]; //这里不要用scanf
        cout<<merge(1,n)<<endl;
    }

    return 0;
}