/*
算法竞赛入门经典177页，bfs，裸题
本题要点：
1、马走日，有8个方向，写好8 个方向， for(int i = 0; i < 8; ++i)
2、套用 bfs 的模型即可。
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
const int MaxN = 9;
bool vis[MaxN][MaxN];
int dx[8] = {
-1, -2, -2, -1, 1, 2, 2, 1};
int dy[8] = {
2, 1, -1, -2, -2, -1, 1, 2};
char s1[3], s2[3];struct node
{
int x, y, cnt;
};node s, t;void bfs()
{
memset(vis, false, sizeof vis); vis[s.x][s.y] = true;queue<node> q;q.push(s);node a, b;while(q.size()){
a = q.front(); q.pop();if(a.x == t.x && a.y == t.y){
printf("To get from %s to %s takes %d knight moves.\n", s1, s2, a.cnt);return;}for(int i = 0; i < 8; ++i){
b.x = a.x + dx[i], b.y = a.y + dy[i], b.cnt = a.cnt + 1;if(b.x >= 0 && b.x < 8 && b.y >= 0 && b.y < 8 && vis[b.x][b.y] == false){
vis[b.x][b.y] = true;q.push(b);}}}
}int main()
{
while(scanf("%s%s", s1, s2) != EOF){
s.x = s1[0] - 'a', s.y = s1[1] - '1', s.cnt = 0; t.x = s2[0] - 'a', t.y = s2[1] - '1'; bfs();}return 0;
}/* e2 e4 a1 b2 b2 c3 a1 h8 a1 h7 h8 a1 b1 c3 f6 f6 *//* To get from e2 to e4 takes 2 knight moves. To get from a1 to b2 takes 4 knight moves. To get from b2 to c3 takes 2 knight moves. To get from a1 to h8 takes 6 knight moves. To get from a1 to h7 takes 5 knight moves. To get from h8 to a1 takes 6 knight moves. To get from b1 to c3 takes 1 knight moves. To get from f6 to f6 takes 0 knight moves. */