題目：

The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.

Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.

1033
1733
3733
3739
3779
8779
8179

The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.

Input

One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).

Output

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

Sample Input

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

題意：

      問從a最少需要改變幾次才能變成b，每次只能改變一位數字，且改變後的數字必須為素數。

題解：

     用廣搜來解決，每次都從千位到各位依次改變一位判斷是否滿足題目要求，滿足即加入佇列。

程式碼：

#include <iostream>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <string.h>
using namespace std;
int a,b;
int turn[4];          //將每次從佇列中取出的數轉化為陣列的形式，便於改變每一位的數字
bool isprime[10003];   //打表，將素數記錄下來
int visited[10000];   //判斷某一數字是否出現過
int step[10000];      //判斷每一位數字從初始狀態到此翻轉了幾次
queue<int>q;

void prime()
{
    int i,j;
	for(i=1000;i<=10000;i++){
		for(j=2;j<i;j++) 
                    if(i%j==0)
                        {isprime[i]=false;break;}
		if(j==i) isprime[i]=true;
	}
}

void turned(int u)
{
    int i;
    for(i=3;i>=0;i--)
    {
        turn[i]=u%10;
        u/=10;
    }
}

int bfs()
{
    int u,i,j;
    q.push(a);
    visited[a]=1;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        if(u==b) return step[u];
        turned(u);          //將數字轉換成陣列形式
        for(i=0;i<4;i++)    //依次遍歷千百十個位
        {
            int x=turn[i];   //x用來還原陣列的數
            for(j=0;j<=9;j++)
            {
                if(i==0&&j==0) continue;
                if(j==x) continue;
                turn[i]=j;
                int v=turn[0]*1000+turn[1]*100+turn[2]*10+turn[3];  //v表示轉換後的數
                if(isprime[v]&&!visited[v])
                {
                    step[v]=step[u]+1;
                    visited[v]=1;
                    q.push(v);
                    if(v==b) return step[v];
                }
                turn[i]=x;    //還原
            }
        }
    }
    return -1;
}

int main()
{
    int n;
    prime();
    cin>>n;
    while(n--)
    {
        cin>>a>>b;
        memset(turn,0,sizeof(turn));
        memset(visited,0,sizeof(visited));
        memset(step,0,sizeof(step));
        while(!q.empty())
            q.pop();
        int ans=bfs();
        if(ans==-1) printf("Impossible\n");
        else cout<<ans<<endl;
    }
    return 0;
}