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.
#!/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:
        ans.append(int(n))
    return list(set(ans))


def greatest_common_divisor(a: int, b: int) -> int:
    """
    Calculate Greatest Common Divisor (GCD).
    >>> greatest_common_divisor(24, 40)
    8
    >>> greatest_common_divisor(1, 1)
    1
    >>> greatest_common_divisor(1, 800)
    1
    >>> greatest_common_divisor(11, 37)
    1
    >>> greatest_common_divisor(3, 5)
    1
    >>> greatest_common_divisor(16, 4)
    4
    >>> greatest_common_divisor(-3, 9)
    3
    >>> greatest_common_divisor(9, -3)
    3
    >>> greatest_common_divisor(3, -9)
    3
    >>> greatest_common_divisor(-3, -9)
    3
    """
    return abs(b) if a == 0 else greatest_common_divisor(b % a, a)


def solveb(n, tuparr):
    pset = set(primeFactors(tuparr[0][0])) | set(primeFactors(tuparr[0][1]))

    for p in pset:
        cnt = 0
        for i, tup in enumerate(tuparr):
            if greatest_common_divisor(tup[0], p) == 1 and greatest_common_divisor(tup[1], p) == 1:
                break
            else:
                cnt += 1
        if cnt == n:
            return p
    return -1


def solve(n, arr):
    m = defaultdict(int)
    for c in arr:
        m[c] += 1
    if max(m.values()) <= 1 and n != 1:
        print('No')
    else:
        print('Yes')


@bootstrap
def rec(i, k, arr, mp, f=0):
    # return str((arr[n - 1] - arr[0]) * (arr[-1] - arr[n]))
    pass


def main():
    # n, k = map(int, input().strip().split())
    # a = []
    # for _ in range(n):
    #     a.append(input().strip())
    # print(solve(n, k, a))
    # for _ in range(int(input().strip())):
    n = int(input().strip())
    # [n, m] = list(map(int, input().strip().split()))
    # arrn = list(map(int, input().strip().split()))
    arr = []
    for _ in range(n):
        arr.append(list(map(int, input().strip().split())))

    print(solveb(n, arr))


######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192


class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0, 2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO, self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill();
            self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s: self.buffer.write(s.encode('ascii'))
            self.read = lambda: self.buffer.read().decode('ascii')
            self.readline = lambda: self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()

# Cout implemented in Python
import sys


class ostream:
    def __lshift__(self, a):
        sys.stdout.write(str(a))
        return self


cout = ostream()
endl = '\n'


# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero=0):
    conv = ord if py2 else lambda x: x
    A = [];
    numb = zero;
    sign = 1;
    i = 0;
    s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0'[0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-'[0]:
                sign = -1
            elif s[i] != b'\r'[0]:
                A.append(sign * numb)
                numb = zero;
                sign = 1
            i += 1
    except:
        pass
    if s and s[-1] >= b'0'[0]:
        A.append(sign * numb)
    return A


if __name__ == "__main__":
    #for _ in range(3):  # todo for testing else remove for loop
        main()
