G

In this problem, you are given an integer number s. You can transform any integer number A to another integer number B by adding x to A. This x is an integer number which is a prime factor of A (please note that 1 and A are not being considered as a factor of A). Now, your task is to find the minimum number of transformations required to transform s to another integer number t.

Input

Input starts with an integer T (≤ 500), denoting the number of test cases.

Each case contains two integers: s (1 ≤ s ≤ 100) and t (1 ≤ t ≤ 1000).

Output

For each case, print the case number and the minimum number of transformations needed. If it's impossible, then print -1.

Sample Input

2

6 12

6 13

Sample Output

Case 1: 2

Case 2: -1

AC代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<queue>
using namespace std;

int c[1010], vis[1010];
int a, b, f;

struct note
{
    int x, s;
};

int prime()
{
    memset(c, -1, sizeof(c));
    c[0] = 0, c[1] = 0;
    for(int i = 2; i*i < 1005; i++)
    {
        if(c[i])
            for(int j = i+i; j < 1005; j += i)
                c[j] = 0;
    }
}

void bfs(int x)
{
    queue<note>Q;
    note p,q;
    p.x = x;
    p.s = 0;
    vis[x] = 1;
    Q.push(p);
    while(!Q.empty())
    {
        p = Q.front();
        Q.pop();
        if(p.x == b)
        {
            f = p.s;
            break;
        }
        for(int i = 2; i < p.x; i++)
        {
            if(c[i] == 0 || p.x + i > b || p.x % i != 0 || vis[p.x+i])
                continue;
            q.x = p.x + i;
            vis[q.x] = 1;
            q.s = p.s + 1;
            Q.push(q);
        }
    }
    return ;
}

int main()
{
    int k = 0, flag = 0;
    prime();

    int t;
    cin >> t;
    while(t--)
    {
        f = -1;
        memset(vis, 0, sizeof(vis));
        cin >> a >> b;
        bfs(a);
        cout << "Case " << ++flag << ": " << f << endl;
    }

    return 0;
}