【Description】

Amakusa, the evil spiritual leader has captured the beautiful princess Nakururu. 
The reason behind this is he had a little problem with Hanzo Hattori, the best 
ninja and the love of Nakururu. After hearing the news Hanzo got extremely angry. 
But he is clever and smart, so, he kept himself cool and
 
 made a plan to face 
Amakusa.

Before reaching Amakusa's castle, Hanzo has to pass some territories. The 
territories are numbered as a, a+1, a+2, a+3 ... b. But not all the territories are
safe for Hanzo because there can be other fighters waiting for him. Actually he is 
not afraid of them, but as
 
 he is facing Amakusa, he has to save his stamina as much 
as possible.

He calculated that the territories which are primes are safe for him. Now given a 
and b he needs to know how many territories are safe for him. But he is busy with 
other plans, so he hired you to solve this small problem!

【Input】

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains a line containing two integers a and b (1 ≤ a ≤ b < 2^31, b - a 
≤ 100000).

【Output】

 For each case, print the case number and the number of safe territories.

【Examples】

Sample Input

3
2 36
3 73
3 11

Sample Output

Case 1: 11
Case 2: 20
Case 3: 4

【Problem Description】

求[a,b]內素數的個數

【Solution】

數論，二次篩
先將sqrt(b)內的素數篩選出來，然後用這部分素數去篩[a,b]內的素數。
可行的原因就在於b-a<=1e5
最後注意一下，如果a==1，答案要減1，因為1不是素數。

【Code】

/*
 * @Author: Simon 
 * @Date: 2018-09-12 13:06:06 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-09-12 14:46:51
 */
#include <bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 100005
int prime[maxn];
bool vis[maxn]={1,1};
int cnt;
void get_prime()
{
    for (int i = 2; i <= maxn; i++)
    {
        if (!vis[i]) prime[++cnt] = i;
        for (int j = 1; j <= cnt && i * prime[j] <= maxn; j++)
        {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) break;
        }
    }
}
int solve(int a, int b)
{
    int vis[maxn] = {0}, ans = 0;
    for (int i = 1; prime[i] * prime[i] <= b; i++)
    {
        int t = (a + prime[i] - 1) / prime[i];
        if (t < 2) t = 2;
        while (t * prime[i] <= b)
        {
            vis[t*prime[i] - a] = 1;
            t++;
        }
    }
    for (int i = 0; i <= b - a; i++) ans += !vis[i];
    return ans;
}
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    get_prime();
    int t;
    cin >> t;
    for (int ca = 1; ca <= t; ca++)
    {
        cout << "Case " << ca << ": ";
        int a, b; cin >> a >> b;
        int ans = solve(a, b);
        cout << ans-(a==1?1:0) << endl;
    }
    cin.get(), cin.get();
    return 0;
}