#!/usr/bin/env python3

# This fails because if the mill are super-efficient,
# the binary search is never entered (since the processing
# time is less than the float_tolerance).
# Consequently, mid never gets a value, and pandemonium ensues


n, w = map(int, input().split())
mills = [tuple(map(int, input().split())) for _ in range(n)]
lo = 2
# upper bound: send everything to the first mill
hi = 2 * mills[0][1] + w / mills[0][0]
while hi - lo > 10**(-6):
    mid = (lo + hi) / 2
    if sum(proces * (mid - 2 * travel) for proces, travel in mills if 2 * travel < mid) >= w:
        hi = mid
    else:
        lo = mid
print(mid)