from lib import *

row = open(0).read().strip()

def is_trap(s):
    assert len(s) == 3
    match s:
        case "^^.":
            return "^"
        case ".^^":
            return "^"
        case "^..":
            return "^"
        case "..^":
            return "^"
        case _:
            return "."

# 400_000 is still easily bruteforcible. If it was a much larget number, we
# would have to cache the rows till we get a repeated row. We would then
# memorize the number of safe tiles from repeated row to repeated row and jump
# forward by multiples of that amount to reach much higher numbers.
r = 0
for _ in range(400_000):
    r += row.count(".")
    nrow = is_trap("." + row[:2])
    for i in range(1, len(row) - 1):
        nrow += is_trap(row[i-1:i+2])
    nrow += is_trap(row[-2:] + ".")
    row = nrow

print(r)