#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
int moves[1001][1001];
int dir[8][2]={-2,-1,-2,1,-1,2,1,2,2,1,2,-1,1,-2,-1,-2};
int b1, b2;
// F = G + H
// G = 从起点到该节点路径消耗
// H = 该节点到终点的预估消耗
struct Knight{
int x,y;
int g,h,f;
bool operator < (const Knight & k) const{ // 重载运算符， 从小到大排序
return < f;
}
};
priority_queue<Knight> que;
int Heuristic(const Knight& k) { // 欧拉距离
return ( - b1) * ( - b1) + ( - b2) * ( - b2); // 统一不开根号，这样可以提高精度
}
void astar(const Knight& k)
{
Knight cur, next;
(k);
while(!())
{
cur=que.top(); ();
if( == b1 && == b2)
break;
for(int i = 0; i < 8; i++)
{
next.x = + dir[i][0];
next.y = + dir[i][1];
if(next.x < 1 || next.x > 1000 || next.y < 1 || next.y > 1000)
continue;
if(!moves[next.x][next.y])
{
moves[next.x][next.y] = moves[][] + 1;
// 开始计算F
next.g = + 5; // 统一不开根号，这样可以提高精度，马走日，1 * 1 + 2 * 2 = 5
next.h = Heuristic(next);
next.f = next.g + next.h;
(next);
}
}
}
}
int main()
{
int n, a1, a2;
cin >> n;
while (n--) {
cin >> a1 >> a2 >> b1 >> b2;
memset(moves,0,sizeof(moves));
Knight start;
start.x = a1;
start.y = a2;
start.g = 0;
start.h = Heuristic(start);
start.f = start.g + start.h;
astar(start);
while(!()) (); // 队列清空
cout << moves[b1][b2] << endl;
}
return 0;
}