#!/usr/bin/env python3
# In the "non-linear fuel cost" situation, we suddenly know
# the median is definitely not optimal. Consider 1,2,100. A
# median of 2 means you'd incur a huge penalty (100 + 99..)
# for the 100 term, 4851 to be exact (plus 1 more for the 1)
#
# A better solution is to optimize for the least number of
# far-out points by using the average. We see 34 is would
# give us 3300 instead, which is significantly better.
import sys


def cost(a, b):
    # Distance 3 = 1 + 2 + 3 = sum(range(abs)))
    return sum(range(abs(a - b) + 1))


crabs = []
for l in sys.stdin:
    crabs = [int(x) for x in l.split(",")]

avg = round(sum(crabs) / len(crabs))

print(sum([cost(x, avg) for x in crabs]))