Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long int mod = 1e9 + 7;
const long long int pinf = ((long long int)2e18);
const long long int ninf = ((long long int)-2e18);
const long long int mod2 = 998244353;
long long int pow(long long int x, unsigned long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int _pow(long long int a, long long int b) {
  if (!b) return 1;
  long long int temp = _pow(a, b / 2);
  temp = (temp * temp);
  if (b % 2) return (a * temp);
  return temp;
}
long long int invmod(long long int a, long long int m) {
  return pow(a, m - 2, m);
}
long long int cl(long long int a, long long int x) {
  if (a % x == 0)
    return a / x;
  else
    return a / x + 1;
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool prime[100000 + 5];
void SieveOfEratosthenes(long long int n) {
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  {
    long long int I, w, h, b, d, m, x, pos, sum = 0, n, k, y, i, j, flag = 1;
    cin >> n;
    vector<long long int> ans;
    for (long long int i = 0; i < n; i++) {
      long long int a, b;
      cin >> a >> b;
      if (i == 0) {
        for (long long int i = 2; i < sqrt(a) + 1; i++) {
          if (a % i == 0) {
            if (isPrime(i)) ans.push_back(i);
            if (a != i * i && (a != i)) {
              if (isPrime(a / i)) ans.push_back(a / i);
            }
          }
        }
        ans.push_back(a);
        if (a != b) {
          for (long long int i = 2; i < sqrt(b) + 1; i++) {
            if (b % i == 0) {
              if (isPrime(i)) ans.push_back(i);
              if (b != i * i && (b != i)) {
                if (isPrime(b / i)) ans.push_back(b / i);
              }
            }
          }
          ans.push_back(b);
        }
      } else {
        for (long long int i = 0; i < ans.size(); i++) {
          if (a % ans[i] != 0 && b % ans[i] != 0) {
            ans.erase(ans.begin() + i);
            i--;
          }
        }
      }
    }
    if (ans.size() == 0) {
      cout << -1;
    } else {
      cout << ans[0];
    }
  }
  return 0;
}