Optimal way of counting the number of non-overlapping pairs given a list of intervals

I'm trying to count the number of non-overlapping pairs given a list of intervals.

For example:

[(1, 8), (7, 9), (3, 10), (7, 12), (11, 13), (13, 14), (9, 15)]

There are 8 pairs:

((1, 8), (11, 13))((1, 8), (13, 14))((1, 8), (9, 15))((7, 9), (11, 13))((7, 9), (13, 14))((3, 10), (11, 13))((3, 10), (13, 14))((7, 12), (13, 14))

I can't seem to figure out a better solution other than to just brute force it by comparing everything with virtually everything else, resulting in a O(n^2) solution.

def count_non_overlapping_pairs(intervals):    intervals = list(set(intervals))  # deduplicate any intervals    intervals.sort(key=lambda x: x[1])    pairs = 0    for i in range(len(intervals)):        for j in range(i+1, len(intervals)):            if intervals[i][1] < intervals[j][0]:                pairs += 1    return pairs

Is there a more optimal solution than this?