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:

 1 2 3 4  3 1033 8179 1373 8017 1033 1033 

output:

 1 2 3  6 7 0 

## Solution

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74  #include #include #include #include using namespace std; // based on the fact that: // if a number n, n!=(6x+1) && n!=(6x+5), // then n cannot be a prime // so we can only check (6x+1) and (6x+5) // while searching for factor, only consider these prime proposals // proof: // n can only \in {6x, 6x+2, 6x+3, 6x+4} // but they are all not prime bool is_prime(int n){ if(n==2 || n==3) return 1; if(n%6!=1 && n%6!=5) return 0; int n_sqrt=floor(sqrt((float)n)); for(int i=5;i<=n_sqrt;i+=6) if(n%(i)==0 | n%(i+2)==0) return 0; return 1; } // return power of 10 int pow_10[5]={1,10,100,1000,10000}; struct node{ int num; int depth; }; int BFS(int first, int last){ bool visit[9000]; memset(visit,0,sizeof(visit)); queue q; while(!q.empty()) q.pop(); node node0={first,0}; q.push(node0); while(!q.empty()){ node node1=q.front(); q.pop(); if(node1.num==last) return node1.depth; for(int i=0; i<4; ++i){ // which bit will be replaced int bit=node1.num%pow_10[4-i]/pow_10[3-i]; for(int j=0; j<10; j+=1){ // first bit cannot be 0 if(i==0 && j==0) continue; // itself if(bit==j) continue; // after replace int num_new=node1.num+((j-bit)*pow_10[3-i]); if(num_new==last) return node1.depth+1; if(visit[num_new-1000]==0 && is_prime(num_new)) q.push({num_new,node1.depth+1}); visit[num_new-1000]=1; } } } return -1; } int main(){ int N,first,last; char s_first[5],s_last[5]; while(scanf("%d",&N)!=EOF){ for(int i=0; i