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

Translation

给两个四位数质数a,ba,b，要求每一次操作改变aa中四位数字的一个数，要求改变后仍为质数，且第一位不能改成00，问至少进行几次操作将aa改成bb。

Idea

每一次操作选择四位数字中的一位改成[0,9][0,9]之间的整数，由于第一位不能是00，因此每次操作共有C41×(9−0+1)−1=39C_4^1\times (9-0+1)-1=39种。
用BFSBFS枚举每一次操作的情况，判断操作后的aa是否为质数（预处理[0,10000][0,10000]之间的质数），并且是否在之前的操作中出现过。

Code

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#include<queue>
#include<cstring>
#include<iostream>
#include<cstdio> 
using namespace std;
const int N=10004;
bool isprime[N];
int prime[N>>1];
int cnt=0;
struct state
{
	//记录一个状态
	int val;//该状态下的a值
	int step;//该状态下的步数
};
void OlaSieve()//欧拉筛预处理质数
{
	memset(isprime,true,sizeof(isprime));
	int i,j;
	for(i=2;i<N;i++)
	{
		if(isprime[i]) prime[++cnt]=i;
		for(j=1;j<=cnt&&i*prime[j]<N;j++)
		{
			isprime[i*prime[j]]=false;
			if(i%prime[j]==0) break;
		}
	}
}
queue<state>q;
int a,b;
bool flag;
bool vis[N];
void init()//初始化
{
	flag=0;
	memset(vis,0,sizeof(vis));
	queue<state>e;
	swap(e,q);
}
void check(state x)//判断新情况是否入队
{
	if(vis[x.val]||!isprime[x.val]) return;
	else
	{
		if(x.val==b) 
		{
			cout<<x.step<<endl;
			flag=1;//到达终点标志
			return;
		}
		else
		{
			vis[x.val]=1;
			q.push(x);
		}
	}
}
void bfs()
{
	//初始状态入队
	state start;
	start.step=0;
	start.val=a;
	q.push(start);
	while(!q.empty())
	{
		state now=q.front();
		state nxt;
		nxt.step=now.step+1;
		int i,j;
		//枚举四位
		for(i=1;i<=4;i++)
		{
			//lp_q表示第p位到第q位的数字
			if(i==1)//更改第一位
			{
				int l2_4=now.val%1000;
				for(j=1;j<=9;j++)
				{
					nxt.val=l2_4+j*1000;
					check(nxt);
					if(flag) return;
				}
			}
			else if(i==2)//更改第二位
			{
				int l1=now.val/1000;
				int l3_4=now.val%100;
				for(j=0;j<=9;j++)
				{
					nxt.val=l1*1000+j*100+l3_4;
					check(nxt);
					if(flag) return;
				}
			}
			else if(i==3)//更改第三位
			{
				int l1_2=now.val/100;
				int l_4=now.val%10;
				for(j=0;j<=9;j++)
				{
					nxt.val=l1_2*100+j*10+l_4;
					check(nxt);
					if(flag) return; 
				}
			}
			else if(i==4)//更改第四位
			{
				int l1_3=now.val/10;
				for(j=0;j<=9;j++)
				{
					nxt.val=l1_3*10+j;
					check(nxt);
					if(flag) return;
				}
			}
		}
		q.pop();//出队
	}
}
void solve()
{
	init();
	cin>>a>>b;
	if(a==b) puts("0");//特判
	else bfs();
}
int main()
{
	OlaSieve();
	int t;
	cin>>t;
	while(t--) solve();
	return 0;
}