## sicily 1444 Prime Path 原

Ciel

### Description

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``````

``````// Problem#: 1444
// Submission#: 1858029
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;

#define MAX 10000
#define DEBUG 1

bool isprime[MAX];
bool visit[MAX];
struct node{
int value;
int step;
node( int a, int b ){
value = a;
step = b;
}
};
int pos[5] = {1,10,100,1000,10000};

void sieve(){
memset(isprime,true,sizeof(isprime));
isprime[0] = isprime[1] = false;
for( int i=2 ; i<=100 ; i++ )
for( int j=i ; i*j<=MAX ; j++ )
if( isprime[i] ) isprime[i*j] = false;
}

inline int digit( int a, int n ){
return (a % pos[n]) / pos[n-1];
}

void bfs( int a, int b ){
queue<node> buffer;
buffer.push(node(a,0));
memset(visit,false,sizeof(visit));
visit[a] = true;
while( !buffer.empty() ){
node tmp = buffer.front();
buffer.pop();
if( tmp.value==b ){
cout << tmp.step << endl;
return ;
}
for( int i=3 ; i>=0 ; i-- ){
node next = tmp;
next.step++;
int temp = next.value - digit(next.value,i+1) * pos[i];
for( int j=0 ; j <= 9 ; j++ ){
next.value = temp + j*pos[i];
if( next.value==tmp.value || next.value<1000 ) continue;
if( isprime[next.value] && !visit[next.value] ){
buffer.push(next);
visit[next.value] = true;
}
}
}
}
cout << "Impossible" << endl;
}

int main(){
sieve();
int n,a,b;
cin >> n;
while(n--){
cin >> a >> b;
bfs(a,b);
}
return 0;
}``````

### Ciel

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