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：求从一个四位数素数到另一个四位数素数所需的最短步数，每一步只能改变一位数字，且改变过程中的数必须是四位数
这是一道典型的数位搜索题，求最短路所以用bfs，难点在于模拟数的转化，另外注意边界控制

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>

using namespace std;

const int maxn=10000;

int n,m;
bool vis[maxn];
int ans;
struct node
{
    int num,step;
};

int mypow(int n,int k)
{
    int res=1;
    while(k--) res*=n;
    return res;
}

bool isprime(int n) //判断素数，由于数据不大，懒得打印素数表了
{
    for(int i=2;i*i<=n;i++){
        if(n%i==0) return false;
    }
    return true;
}

int change(int n,int i,int j) //模拟数的转化，有很多技巧方法，一定要细心
{
    int f=n/mypow(10,i+1)*mypow(10,i+1);//保留前缀
    int r=n%mypow(10,i);//保留后缀
    int mid=j*mypow(10,i); //改变中间的数
    return f+mid+r;
}

bool bfs()
{
    memset(vis,0,sizeof(vis));
    queue<node> q;
    q.push({n,0});
    vis[n]=1;
    while(!q.empty()){
        node now=q.front();
        q.pop();
        if(now.num==m){
            ans=now.step;
            return true;
        }
        for(int i=0;i<4;i++){  //从第1位到第四位
            for(int j=0;j<=9;j++){  //把所在位数字改为j
                int nextnum=change(now.num,i,j);
                if(nextnum<1000||nextnum>=maxn) continue;//边界控制
                if(vis[nextnum]||!isprime(nextnum)) continue;
                vis[nextnum]=1;
                q.push({nextnum,now.step+1});
            }
        }
    }
    return false;
}

int main()
{
    int T;cin>>T;
    while(T--){
        cin>>n>>m;
        if(bfs()) cout<<ans<<endl;
        else cout<<"Impossible"<<endl;
    }
    return 0;
}