最大公约数_1

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 andb 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 integern, 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.

Sample Input

Input

9 2731 510 119 11

Output

3-19

题意：给定一个范围（a,b）和两个数，然后就是找出两个数在这个范围内的最大公约数

解题技巧：先找出最短公约数，然后在找最大公约数的约数
    利用upper_bound找出边界

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include"algorithm"using namespace std;int j,n,m;int a[1000000];int GCD(int a,int b){    if (a % b == 0)        return b;    return GCD(b , a % b);}int main(){    while(~scanf("%d%d",&n,&m))    {        /*预处理*/        j=0;        memset(a,0,sizeof(a));        int g = GCD (n,m);        for(int i=1;i<=sqrt((double)(g));i++)        {                if (g % i == 0)                {                    a[j++] = i;                    a[j++] = g / i;                }            }        sort(a,a+j);        int n;        int l,c,r;        scanf("%d",&n);        for(int i=0;i<n;i++)        {            scanf ("%d %d",&l,&r);            int t = upper_bound(a , a + j , r) - a -1;        //返回的是大于它的地址，首先把这个地址减一，然后减去首地址得到下标            if (a[t] < l)                printf ("-1\n");            else                printf ("%d\n",a[t]);        }    }    return 0;}