import sys

sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]

def main():
    def isl(m):
        mx0 = mx1 = 0
        mn0 = mn1 = inf
        for a, b in zip(aa, bb):
            v = a + b * (m - 1)
            if v > mx0: mx0 = v
            if v < mn0: mn0 = v
            v = a + b * m
            if v > mx1: mx1 = v
            if v < mn1: mn1 = v
        return mx0 - mn0 > mx1 - mn1

    def f(x):
        mx = 0
        mn = inf
        for a, b in zip(aa, bb):
            v = a + b * x
            if v > mx: mx = v
            if v < mn: mn = v
        return mx - mn

    inf = 10 ** 19
    n = II()
    aa = []
    bb = []
    for _ in range(n):
        a, b = MI()
        aa.append(a)
        bb.append(b)
    # y=a+bxとして、f(x)=max(y)-min(y)は下に凸の関数になる
    # 下に凸の関数の最小値は三分探索が一般的だが、今回はxが整数という制約があるので二分探索でいける
    # あるx1と、その左隣のx0（つまりx0=x1-1）があったとき
    # f(x0)>f(x1)となればx1は最小値を含む区間の左端になる
    # そうでなければx1は右端
    l = 0
    r = 10**9+1
    while l + 1 < r:
        m = (l + r) // 2
        if isl(m): l = m
        else: r = m
    # 一応lrの前後を含めて確認
    mn = inf
    ans = r
    for x in range(max(1, l - 1), r + 2):
        v = f(x)
        if v < mn:
            ans = x
            mn = v
    print(ans)

main()