#include<bits/stdc++.h>
#define MAXN 1000005
using namespace std;
typedef long long ll;

ll prime[MAXN];
ll vis[MAXN];
ll cnt;
ll n;
void isprime()
{
    cnt=0;
    for(int i=2;i<=MAXN;i++)
    {
        if(!vis[i])prime[cnt++]=i;
        for(int j=0;j<cnt && i*prime[j]<=MAXN;j++)
        {
            vis[i*prime[j]]=i;
            if(i%prime[j]==0)break;
        }
    }
}
ll solve()
{
    ll sum=n;
    ll ans=1;
    for(int i=0;i<cnt && prime[i]<sum;i++)
    {
        int k=0;
        while(sum%prime[i]==0)
        {
            sum/=prime[i];
            k++;
        }
        ans*=(1+k);
    }
    if(sum>1)ans*=(1+1);//多出一個素數來
    return ans;
}
 

                             Aladdin and the Flying Carpet

It's said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concerned about the first mystery.

Aladdin was about to enter to a magical cave, led by the evil sorcerer who disguised himself as Aladdin's uncle, found a strange magical flying carpet at the entrance. There were some strange creatures guarding the entrance of the cave. Aladdin could run, but he knew that there was a high chance of getting caught. So, he decided to use the magical flying carpet. The carpet was rectangular shaped, but not square shaped. Aladdin took the carpet and with the help of it he passed the entrance.

Now you are given the area of the carpet and the length of the minimum possible side of the carpet, your task is to find how many types of carpets are possible. For example, the area of the carpet 12, and the minimum possible side of the carpet is 2, then there can be two types of carpets and their sides are: {2, 6} and {3, 4}.

Input

Input starts with an integer T (≤ 4000), denoting the number of test cases.

Each case starts with a line containing two integers: a b (1 ≤ b ≤ a ≤ 1012) where a denotes the area of the carpet and b denotes the minimum possible side of the carpet.

Output

For each case, print the case number and the number of possible carpets.

Sample Input

2

10 2

12 2

Sample Output

Case 1: 1

Case 2: 2

題意：

給你一個長方形面積和一個最小可能的邊長，統計有多少種滿足面積相等且邊長大於等於最小邊長的長方形（一定不是正方形）。

根據唯一分解定理，先將a唯一分解，則a的所有正約數的個數為num = (1 + a1) * (1 + a2) *...(1 + ai)，這裡的ai是素因子的指數，見唯一分解定理，因為題目說了不會存在c==d的情況，因此num要除2，去掉一半，然後列舉小於b的a的約數，拿num減掉就可以了

注意當b*b>n 直接不走不然會超時

#include<bits/stdc++.h>
#define MAXN 1000005
using namespace std;
typedef long long ll;

ll prime[MAXN];
ll vis[MAXN];
ll cnt;
ll n;
void isprime()
{
    cnt=0;
    for(int i=2;i<=MAXN;i++)
    {
        if(!vis[i])prime[cnt++]=i;
        for(int j=0;j<cnt && i*prime[j]<=MAXN;j++)
        {
            vis[i*prime[j]]=i;
            if(i%prime[j]==0)break;
        }
    }
}
ll solve()
{
    ll sum=n;
    ll ans=1;
    for(int i=0;i<cnt && prime[i]<sum;i++)
    {
        int k=0;
        while(sum%prime[i]==0)
        {
            sum/=prime[i];
            k++;
        }
        ans*=(1+k);
    }
    if(sum>1)ans*=(1+1);//多出一個素數來
    return ans;
}
int main()
{
    int t;
    int CS=1;
    isprime();
    ll d;
    scanf("%d",&t);
    ll ans;
    while(t--)
    {
        ans=0;
        ll cnt=0;
        scanf("%lld%lld",&n,&d);
        if(d*d>n)
            ans=0;
        else
        {
            for(int i=1;i<d;i++)if(n%i==0)cnt++;
            ans=solve()/2-cnt;
        }

        
        printf("Case %d: %lld\n",CS++,ans);

    }

    return 0;
}