Modified GCD CodeForces - 75C

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;
}