Later days that closed lower

You are given $n$ daily closing prices $p_0, \dots, p_{n-1}$. For each day $i$, return the number of later days $j > i$ with $p_j < p_i$. Checking every pair is $O(n^2)$; do it in $O(n \log n)$. For example, closes $[101, 99, 103, 98, 100]$ give $[3, 1, 2, 0, 0]$.