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

Output

Sample Input

3

1033 8179

1373 8017

1033 1033

Sample Output

6

7

0

给出两个数s，e，都是素数。经过转化，将s转化为e的最小步数

规则，每一次仅仅能改动一位，每次得到的数都是素数。

素数筛跑出1000到10000内的全部素数。假设当中两个素数仅仅有一位不同。那么连接一条边。得到全部素数组合的图后用bfs直接搜索就能够

#include <cstdio>

#include <cstring>

#include <queue>

#include <algorithm>

using namespace std;

int a[11000] , check[11000] , tot ;

struct node{

    int v , next ;

}p[2000000];

struct node1{

    int u , t ;

};

queue <node1> que ;

int head[10000] , cnt , flag[10000] ;

void add(int u,int v)

{

    p[cnt].v = v ;

    p[cnt].next = head[u] ;

    head[u] = cnt++ ;

}

void init()

{

    memset(check,0,sizeof(check));

    memset(head,-1,sizeof(head));

    tot = cnt = 0 ;

    int i , j , k , num ;

    for(i = 2 ; i <= 10000 ; i++)

    {

        if( !check[i] )

            a[tot++] = i ;

        for(j = 0 ; j < tot ; j++)

        {

            if(i*a[j] >= 10000)

                break;

            check[i*a[j]] = 1 ;

            if( i%a[j] == 0 )

                break;

        }

    }

    for(i = 0 ; i < tot ; i++)

        if( (a[i]/1000) ) break;

    k = i ;

    for(i = k ; i < tot ; i++)

    {

        for(j = k ; j < i ; j++)

        {

            num = 0 ;

            if( a[i]%10 != a[j]%10 )

                num++ ;

            if( a[i]/10%10 != a[j]/10%10 )

                num++ ;

            if( a[i]/100%10 != a[j]/100%10 )

                num++ ;

            if( a[i]/1000%10 != a[j]/1000%10 )

                num++ ;

            if(num == 1)

            {

                add(i,j);

                add(j,i);

            }

        }

    }

}

int find1(int x)

{

    int low = 0 , mid , high = tot-1 ;

    while(low <= high)

    {

        mid = (low+high)/2 ;

        if(a[mid] == x)

            return mid ;

        else if(a[mid] < x)

            low = mid + 1 ;

        else

            high = mid -1 ;

    }

}

int bfs(int s,int e)

{

    memset(flag,0,sizeof(flag));

    while( !que.empty() )

        que.pop();

    int i , j , v ;

    node1 low , high ;

    low.u = s ;

    low.t = 0 ;

    flag[s] = 1 ;

    que.push(low);

    while( !que.empty() )

    {

        low = que.front();

        que.pop();

        if( low.u == e )

            return low.t ;

        for(i = head[low.u] ; i != -1 ; i = p[i].next)

        {

            v = p[i].v ;

            if( !flag[v] )

            {

                flag[v] = 1;

                high.u = v ;

                high.t = low.t + 1 ;

                que.push(high);

            }

        }

    }

    return 0;

}

int main()

{

    int t , s , e ;

    init();

    scanf("%d", &t);

    while(t--)

    {

        scanf("%d %d", &s, &e);

        s = find1(s);

        e = find1(e);

        printf("%d\n", bfs(s,e) );

    }

    return 0;

}