CF1025B Weakened Common Divisor 数学

Weakened Common Divisor

time limit per test

1.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 (a1,b1)(a1,b1), (a2,b2)(a2,b2), ..., (an,bn)(an,bn) their WCD is arbitrary integer greater than 11, 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)][(12,15),(25,18),(10,24)], then their WCD can be equal to 22, 33, 55 or 66 (each of these numbers is strictly greater than 11 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 nn (1≤n≤1500001≤n≤150000) — the number of pairs.

Each of the next nn lines contains two integer values aiai, bibi (2≤ai,bi≤2⋅1092≤ai,bi≤2⋅109).

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 −1−1.

Examples

input

Copy

3
17 18
15 24
12 15

output

Copy

6

input

Copy

2
10 16
7 17

output

Copy

-1

input

Copy

5
90 108
45 105
75 40
165 175
33 30

output

Copy

5

Note

In the first example the answer is 66 since it divides 1818 from the first pair, 2424 from the second and 1212 from the third ones. Note that other valid answers will also be accepted.

In the second example there are no integers greater than 11 satisfying the conditions.

In the third example one of the possible answers is 55. Note that, for example, 1515 is also allowed, but it's not necessary to maximize the output.

题意：有n组数，每组数有两个数，求一个数是所有组数中的两个中一个的因子

分析：分解第一组数得到他们的质因子，如果这些数有解，则这些因子肯定有一个是其他所有组数中至少一个数的因子

　　枚举剩下每组数

AC代码：

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const ll mod = 998244353;
const double pi = acos(-1.0);
const double eps = 1e-8;
ll a[maxn], b[maxn];
int main() {
    ios::sync_with_stdio(0);
    ll n, x, y;
    set<ll> s, t;
    cin >> n;
    for( ll i = 0; i < n; i ++ ) {
        cin >> a[i] >> b[i];
    }
    x = a[0], y = b[0];
    for( ll i = 2; i*i <= x; i ++ ) {
        if( x%i == 0 ) {
            s.insert(i);
            while( x%i == 0 ) {
                x /= i;
            }
        }
    }
    for( ll i = 2; i*i <= y; i ++ ) {
        if( y%i == 0 ) {
            s.insert(i);
            while( y%i == 0 ) {
                y /= i;
            }
        }
    }
    if( x > 1 ) {
        s.insert(x);
    }
    if( y > 1 ) {
        s.insert(y);
    }
    bool flg = false;
    for( ll i : s ) {
        bool flag = true;
        for( ll j = 1; j < n; j ++ ) {
            if( a[j]%i && b[j]%i ) {
                flag = false;
                break;
            }
        }
        if(flag) {
            cout << i << endl;
            flg = true;
            break;
        }
    }
    if(!flg) {
        cout << -1 << endl;
    }
    return 0;
}