count = a(i) * (sum - s(i))
#include <iostream>
using namespace std;
const int N = 200010;
typedef long long ll;
int n;
int s[N];
int main()
{
ll sum = 0;
scanf("%d", &n);
for(int i = 1; i <= n; i++)
{
scanf("%d", &s[i]);
sum += s[i];
}
ll cnt = 0, count = 0;
for(int i = 1; i <= n; i++)
{
cnt += s[i];
count += s[i] * (sum - cnt);
}
printf("%lld", count);
return 0;
}