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.

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).

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

3

1033 8179

1373 8017

1033 1033

6

7

0

题意：题目较长，实际就是给你两个四位素数，让你每次只能更改第一个素数的其中一位，更改后要求也是素数且位数不变，问你至少需要更改几次才能变成第二个素数。无解输出Impossible。
思路：本题涉及到素数，每次更改后均需要判断，所以避免重复计算，在程序开始先用筛法把每个四位数的素数性提前存到数组prime。之后分别更改一位数值（第一位不可能是0，最后一位只能是奇数），记录下变更次数即可。

#include<stdio.h>

#include<string.h>

#include<queue>

using namespace std;

int prime[],bo[];

struct Node{

    int x,s;

}node;

int main()

{

    int t,a,b,f,i,j;

    prime[]=;

    for(i=;i<=;i++){

        if(!prime[i]){

            for(j=;i*j<=;j++){

                prime[i*j]=;     //素数筛选

            }

        }

    }

    scanf("%d",&t);

    while(t--){

        queue<Node> q;

        memset(bo,,sizeof(bo));

        scanf("%d%d",&a,&b);

        if(a==b) printf("0\n");

        else{

            bo[a]=;

            node.x=a;

            node.s=;

            q.push(node);

            f=;

            while(q.size()){

                int tx=q.front().x;

                for(i=;i<=;i++){

                    if(i!=tx%&&!prime[tx-tx%+i]&&bo[tx-tx%+i]==){

                        if(tx-tx%+i==b){

                            f=q.front().s+;

                            break;

                        }

                        bo[tx-tx%+i]=;

                        node.x=tx-tx%+i;

                        node.s=q.front().s+;

                        q.push(node);

                    }

                    if(i!=tx/%&&!prime[tx-tx/%*+i*]&&bo[tx-tx/%*+i*]==){

                        if(tx-tx/%*+i*==b){

                            f=q.front().s+;

                            break;

                        }

                        bo[tx-tx/%*+i*]=;

                        node.x=tx-tx/%*+i*;

                        node.s=q.front().s+;

                        q.push(node);

                    }

                    if(i!=tx/%&&!prime[tx-tx/%*+i*]&&bo[tx-tx/%*+i*]==){

                        if(tx-tx/%*+i*==b){

                            f=q.front().s+;

                            break;

                        }

                        bo[tx-tx/%*+i*]=;

                        node.x=tx-tx/%*+i*;

                        node.s=q.front().s+;

                        q.push(node);

                    }

                    if(i!=&&i!=tx/&&!prime[tx-tx/*+i*]&&bo[tx-tx/*+i*]==){

                        if(tx-tx/*+i*==b){

                            f=q.front().s+;

                            break;

                        }

                        bo[tx-tx/*+i*]=;

                        node.x=tx-tx/*+i*;

                        node.s=q.front().s+;

                        q.push(node);

                    }

                }

                if(f!=) break;

                q.pop();

            }

            if(f==) printf("Impossible\n");

            else printf("%d\n",f);

        }

    }

    return ;

}