Algorithm

C. Modified GCD
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem Link- https://codeforces.com/problemset/problem/75/C
input
standard input
output
standard output

Well, here is another math class task. In mathematics, GCD is the greatest common divisor, and it's an easy task to calculate the GCD between two positive integers.

A common divisor for two positive numbers is a number which both numbers are divisible by.

But your teacher wants to give you a harder task, in this task you have to find the greatest common divisor d between two integers a and b that is in a given range from low to high (inclusive), i.e. low ≤ d ≤ high. It is possible that there is no common divisor in the given range.

You will be given the two integers a and b, then n queries. Each query is a range from low to high and you have to answer each query.

Input

The first line contains two integers a and b, the two integers as described above (1 ≤ a, b ≤ 109). The second line contains one integer n, the number of queries (1 ≤ n ≤ 104). Then n lines follow, each line contains one query consisting of two integers, low and high (1 ≤ low ≤ high ≤ 109).

Output

Print n lines. The i-th of them should contain the result of the i-th query in the input. If there is no common divisor in the given range for any query, you should print -1 as a result for this query.

Examples
input
Copy
9 27
3
1 5
10 11
9 11
output
Copy
3
-1
9



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <cstdio>
#include <vector>
#include <algorithm>

long gcd(long a, long b){return (b == 0) ? a : gcd (b, a%b);}

int main(){

    long a, b; scanf("%ld %ld\n", &a, &b);
    long g = gcd(a, b);
    std::vector<long> div;
    for(long p = 1; p * p <= g; p++){
        if(g % p == 0){
            div.push_back(p);
            if(p * p < g){div.push_back(g / p);}
        }
    }
    sort(div.begin(), div.end());

    long q; scanf("%ld", &q);
    while(q--){
        long low, high; scanf("%ld %ld", &low, &high);
        long mgcd = *(--(std::upper_bound(div.begin(), div.end(), high)));
        printf("%ld\n", (mgcd >= low) ? mgcd : -1);
    }

    return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
9 27
3
1 5
10 11
9 11

Output

x
–
+
cmd
3
-1
9
Advertisements

Demonstration

Codeforces Solution-C. Modified GCD-Solution in C, C++, Java, Python

Previous
Codeforces solution 1080-B-B. Margarite and the best present codeforces solution
Next
CodeChef solution DETSCORE - Determine the Score CodeChef solution C,C+