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

 1 //
 2 // Created by w on 2020/11/5.
 3 //
 4 
 5 #include <iostream>
 6 #include <queue>
 7 #include <cstring>
 8 #include <cmath>
 9 using namespace std;
10 typedef long long ll ;
11 const int N=10000;
12 int prime[N];
13 bool visit[N];
14 int cnt;
15 struct st
16 {
 
17     int num;
18     int step;
19 };
20 void Prime()//篩素數,因為題裡是4位數，所以從1000-10000的素數篩了出來
21 {
22     int m = sqrt(N + 0.5);
23     for(int i = 2; i <= m; i++)
24     {
25         if(!visit[i])
26         {
27             for(int j = i * i; j <= N; j += i)
28                 visit[j] = 1;
29         }
30     }
31     for(int i = 1000; i <= N; i++)
32     {
33         if(!visit[i]) prime[cnt++] = i;
34     }
35 }
36 int ck(int a,int b)
37 {
38     int ans=0,x,y;
39     for(int i=0; i<4; i++)
40     {
41         x=a%10;
42         y=b%10;
43         if(x!=y)
44             ans++;//統計不一樣的位數的數字，就是需要購買的
45         a/=10;
46         b/=10;
47     }
48     return ans;
49 }
50 int bfs(int a,int b)
51 {
52     queue<st> q;
53     st a1,b1;
54     a1.num=a;
55     a1.step=0;//step表示步長，需要更換的門牌數字個數
56     q.push(a1);//a1入隊
57     while (!q.empty())
58     {
59         b1=q.front();
60         q.pop();
61         if(b1.num==b)
62             return b1.step;
63         for(int i=0; i<cnt; i++)
64         {
65             if(visit[i])//訪問過的素數跳過
66                 continue;
67             a1.num=prime[i];//是素數儲存到a1.num
68             a1.step=b1.step+1;//步長加1
69             int y=ck(b1.num,a1.num);//比較數字不同的位數
70             if(y>1)//一位以上重複執行此過程統計步長
71                 continue;
72             else
73             {
74                 visit[i]=1;//標記此時這個素數已經被用過
75                 if(y==1)//僅有一位不同，會被重新放入佇列，進行下一次比較
76                     q.push(a1);
77             }
78         }
79     }
80     return -1;//沒找到返回-1
81 
82 }
83 int main()
84 {
85     int t,m,n;
86     Prime();
87     cin>>t;
88     while (t--)
89     {
90         memset(visit,0,sizeof(visit));//重置visit陣列
91         cin>>m>>n;
92         int ans=bfs(m,n);
93         if(ans!=-1)
94             cout<<ans<<endl;
95         else
96             cout<<"Impossible"<<endl;
97     }
98     return  0;
99 }