#!/usr/bin/env python3
import sys
import re

buta_re = re.compile("Button A: X\+(\d+), Y\+(\d+)")
butb_re = re.compile("Button B: X\+(\d+), Y\+(\d+)")
priz_re = re.compile("Prize: X=(\d+), Y=(\d+)")

def parse():
    games = []
    game = {}
    for l in sys.stdin:
        am = buta_re.match(l)
        bm = butb_re.match(l)
        pm = priz_re.match(l)
        if am:
            game["A"] = [int(x) for x in am.groups(1)]
        elif bm:
            game["B"] = [int(x) for x in bm.groups(1)]
        elif pm:
            game["P"] = [int(x) for x in pm.groups(1)]
        else:
            games.append(game)
            game = {}
    if game:
        games.append(game)
    return games

# Its a linear equation, so they should interesect exactly
# once, not at all, or continuously (same line).
# Searching the entire b space first (inner loop) gives
# us the cheapest if they're the same line
#
# For pt 1, we just brute force it rather than solve it
def solve(game):
    for a in range(101):
        for b in range(101):
            if a * game['A'][0] + b * game['B'][0] == game['P'][0]:
                if a * game['A'][1] + b * game['B'][1] == game['P'][1]:
                    return (a, b)
    return None


if __name__ == '__main__':
    tot = 0
    games = parse()
    for g in games:
        sol = solve(g)
        if sol is not None:
            tot += sol[0] * 3 + sol[1]
    print(tot)