Modified GCD CodeForces

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.

Example

Input

9 27
3
1 5
10 11
9 11

Output

3
-1
9

题解：找出a,b的所有公共约数，然后二分查找区间左右端点的位置，分类讨论

#include<bits/stdc++.h>
using namespace std;

const int maxn=2e5+10;

int a,b,n;
int ma[maxn];

int gcd(int a,int b){
    if(b==0) return a;
    return gcd(b,a%b);
}

int main()
{   cin>>a>>b>>n;
    int g=gcd(a,b),cnt=0;
    for(int j=1;j<=sqrt(g)+1;j++){
        if(g%j!=0) continue;
        ma[cnt++]=j;
        if(j<g/j) ma[cnt++]=g/j;
    }
    sort(ma,ma+cnt);
    int aa,bb;
    for(int i=1;i<=n;i++){
        cin>>aa>>bb;
        //for(int j=0;j<cnt;j++) cout<<ma[j]<<endl;
        int pos1=lower_bound(ma,ma+cnt,aa)-ma;
        int pos2=lower_bound(ma,ma+cnt,bb)-ma;
        //cout<<pos1<<" "<<pos2<<endl;
        if(pos1==pos2){
            if(ma[pos2]==bb) cout<<bb<<endl;
            else cout<<"-1"<<endl;
        }
        else{
            if(ma[pos2]==bb) cout<<ma[pos2]<<endl;
            else cout<<ma[pos2-1]<<endl;
        }
    }
    return 0;
}