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

　　题目大意：n组输入，每一组输入两个四位质数，对第一个质数进行转换，使之变为第二个质数，每次只能更改一个数中某一位的数，且变动一位后的数也是质数才符合要求。输出最
少经过几次转换，能转换成第二个质数，保证输入合法。
　　思路：我这里采用了比较笨的方法，用结构体存储每一位上的数，以及总数，建立结构体队列，进行广度优先搜索，通过数组存储转换次数信息。（主要是自己比较菜，获取某个
数中每一位的数不够熟练，后续会补上用整型解决问题的思路。）还有一个问题是判断某数是否为质数的函数，这里我参考了大佬的博客（https://blog.csdn.net/huang_miao_xin/article/details/51331710），
很巧妙的判断质数的算法，值得学习。代码如下：

#include <iostream>
#include <algorithm>
#include <math.h>
#include <queue>
#include <string.h>
int const max_n = 10002;
using namespace std;
int jug[max_n];//记录信息数组
struct node {
    int a, b, c, d;//千位，百位，十位，个位
    int sum;//数的大小
};
queue<node> q;
int test(node x)//由各个数位的数求大小
{
    return x.a * 1000 + x.b * 100 + x.c * 10 + x.d;
}
bool judge(int x)//是否为质数
{//关于这种方法判断质数的解释，6x，6x+2，6x+4一定被2整除，6x+3一定被3整除，这样就只用判断6x+1和6x+5的数了。而当循环以6位单位跨进时，就是6x-1和6x+1的情况了。更详细的讲解可以去大佬博客看
    if (x == 2 || x == 3)return true;//特判2,3两个质数
    if (x % 6 != 1 && x % 6 != 5)return false;
    int tmp = sqrt(x);
    for (int i = 5; i <= tmp; i += 6)
    {
        if (x%i == 0 || x % (i + 2) == 0)return false;
    }
    return true;
}
int bfs(int x, int y)
{
    node sx;
    sx.a = x / 1000;
    sx.b = x / 100 % 10;
    sx.c = x / 10 % 10 % 10;
    sx.d = x % 1000 % 100 % 10;
    sx.sum = x;
    q.push(sx);//入队
    memset(jug, 0, sizeof(jug));
    jug[x] = 1;
    while (!q.empty())
    {
        node s = q.front(); q.pop();
        if (s.sum == y)return jug[s.sum] - 1;
        for (int i = 1; i < 10; i++)//千位进行转换
        {
            node p;
            p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
            p.a = i;
            p.sum = test(p);
            if (judge(p.sum) && jug[p.sum] == 0)
            {
                jug[p.sum] = jug[s.sum] + 1;
                q.push(p);
            }
        }
        for (int i = 0; i < 10; i++)//百位进行转换
        {
            node p;
            p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
            p.b = i;
            p.sum = test(p);
            if (judge(p.sum) && jug[p.sum] == 0)
            {
                jug[p.sum] = jug[s.sum] + 1;
                q.push(p);
            }
        }
        for (int i = 0; i < 10; i++)//十位进行转换
        {
            node p;
            p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
            p.c = i;
            p.sum = test(p);
            if (judge(p.sum) && jug[p.sum] == 0)
            {
                jug[p.sum] = jug[s.sum] + 1;
                q.push(p);
            }
        }
        for (int i = 0; i < 10; i++)//个位进行转换
        {
            node p;
            p.a = s.a, p.b = s.b, p.c = s.c, p.d = s.d;
            p.d = i;
            p.sum = test(p);
            if (judge(p.sum) && jug[p.sum] == 0)
            {
                jug[p.sum] = jug[s.sum] + 1;
                q.push(p);
            }
        
        }
    }
    return 0;
}
int main()
{
    int t; scanf("%d", &t);
    while (t--)
    {
        int m, n;
        scanf("%d %d", &m, &n);
        while (q.size())q.pop();
        if (m == n) { printf("0\n"); continue; }
        else
        {
            int x = bfs(m, n);
            if (x == 0)printf("Impossible\n");
            else printf("%d\n", x);
        }
    }
    return 0;
}