human
During the research on properties of the greatest common divisor (GCD) of a set of numbers, Ildar, a famous mathematician, introduced a brand new concept of the weakened common divisor (WCD) of a list of pairs of integers.

For a given list of pairs of integers (a_1, b_1), (a_2, b_2), ..., (a_n, b_n) their WCD is arbitrary integer greater than 1, such that it divides at least one element in each pair. WCD may not exist for some lists.

For example, if the list looks like [(12, 15), (25, 18), (10, 24)], then their WCD can be equal to 2, 3, 5 or 6 (each of these numbers is strictly greater than 1 and divides at least one number in each pair).

You're currently pursuing your PhD degree under Ildar's mentorship, and that's why this problem was delegated to you. Your task is to calculate WCD efficiently.

Input

The first line contains a single integer n (1 ≤ n ≤ 150 000) — the number of pairs.

Each of the next n lines contains two integer values a_i, b_i (2 ≤ a_i, b_i ≤ 2 ⋅ 10^9).

Output

Print a single integer — the WCD of the set of pairs. 

If there are multiple possible answers, output any.
if there is no answer, print -1.

Examples

Input

3
17 18
15 24
12 15

Output

6

Input

2
10 16
7 17

Output

-1

Input

5
90 108
45 105
75 40
165 175
33 30

Output

5

Note

In the first example the answer is 6 since it divides 18 from the first pair, 24 from the second and 12 from the third ones. Note that other valid answers will also be accepted.

In the second example there are no integers greater than 1 satisfying the conditions.

In the third example one of the possible answers is 5. Note that, for example, 15 is also allowed, but it's not necessary to maximize the output.
Solve the task in PYTHON3.
assistant
#Here is Python code to print a single integer — the WCD of the set of pairs.

#!/usr/bin/env python
import os
import sys

from io import BytesIO, IOBase

# Bootstrap https://github.com/cheran-senthil/PyRival/blob/master/tests/misc/test_bootstrap.py
from types import GeneratorType

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc

#################

from collections import defaultdict
from math import sqrt

INF = pow(2, 63)

def isPrime(number):
    """
    input: positive integer 'number'
    returns true if 'number' is prime otherwise false.
    """

    # precondition
    assert isinstance(number, int) and (
            number >= 0
    ), "'number' must been an int and positive"

    status = True

    # 0 and 1 are none primes.
    if number <= 1:
        status = False

    for divisor in range(2, int(round(sqrt(number))) + 1):

        # if 'number' divisible by 'divisor' then sets 'status'
        # of false and break up the loop.
        if number % divisor == 0:
            status = False
            break

    # precondition
    assert isinstance(status, bool), "'status' must been from type bool"

    return status

import math
def primeFactors(n):
    # Print the number of two's that divide n
    ans = []
    while n % 2 == 0:
        ans.append(2)
        n = n / 2

    # n must be odd at this point
    # so a skip of 2 ( i = i + 2) can be used
    for i in range(3, int(math.sqrt(n)) + 1, 2):

        # while i divides n , print i ad divide n
        while n % i == 0:
            ans.append(int(i))
            n = n / i

            # Condition if n is a prime
    # number greater than 2
    if n > 2:
       ...