Problem:
Score : 600 points

Problem Statement

For a non-negative integer K, we define a fractal of level K as follows:

A fractal of level 0 is a grid with just one white square.

When K > 0, a fractal of level K is a 3^K \times 3^K grid. If we divide this grid into nine 3^{K-1} \times 3^{K-1} subgrids:

The central subgrid consists of only black squares.

Each of the other eight subgrids is a fractal of level K-1.

For example, a fractal of level 2 is as follows:

In a fractal of level 30, let (r, c) denote the square at the r-th row from the top and the c-th column from the left.

You are given Q quadruples of integers (a_i, b_i, c_i, d_i).
For each quadruple, find the distance from (a_i, b_i) to (c_i, d_i).

Here the distance from (a, b) to (c, d) is the minimum integer n that satisfies the following condition:

There exists a sequence of white squares (x_0, y_0), \ldots, (x_n, y_n) satisfying the following conditions:

(x_0, y_0) = (a, b)

(x_n, y_n) = (c, d)

For every i (0 \leq i \leq n-1), (x_i, y_i) and (x_{i+1}, y_{i+1}) share a side.

Constraints

1 \leq Q \leq 10000

1 \leq a_i, b_i, c_i, d_i \leq 3^{30}

(a_i, b_i) \neq (c_i, d_i)

(a_i, b_i) and (c_i, d_i) are white squares.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

Q
a_1 \ b_1 \ c_1 \ d_1
:
a_Q \ b_Q \ c_Q \ d_Q

Output

Print Q lines.
The i-th line should contain the distance from (a_i, b_i) to (c_i, d_i).

Sample Input 1

2
4 2 7 4
9 9 1 9

Sample Output 1

5
8

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
#include<cassert>
#include<complex>
#include<numeric>
using namespace std;

//#define int long long
typedef long long ll;

typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
const ll INF = mod * mod;
typedef pair<int, int>P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
#define all(v) (v).begin(),(v).end()
typedef pair<ll, ll> LP;
typedef long double ld;
typedef pair<ld, ld> LDP;
const ld eps = 1e-12;
const ld pi = acos(-1.0);
//typedef vector<vector<ll>> mat;
typedef vector<int> vec;

ll mod_pow(ll a, ll n, ll m) {
	ll res = 1;
	while (n) {
		if (n & 1)res = res * a%m;
		a = a * a%m; n >>= 1;
	}
	return res;
}

struct modint {
	ll n;
	modint() :n(0) { ; }
	modint(ll m) :n(m) {
		if (n >= mod)n %= mod;
		else if (n < 0)n = (n%mod + mod) % mod;
	}
	operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
modint operator+=(modint &a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= mod; return a; }
modint operator-=(modint &a, modint b) { a.n -= b.n; if (a.n < 0)a.n += mod; return a; }
modint operator*=(modint &a, modint b) { a.n = ((ll)a.n*b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, int n) {
	if (n == 0)return modint(1);
	modint res = (a*a) ^ (n / 2);
	if (n % 2)res = res * a;
	return res;
}

ll inv(ll a, ll p) {
	return (a == 1 ? 1 : (1 - p * inv(p%a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }

const int max_n = 1 << 18;
modint fact[max_n], factinv[max_n];
void init_f() {
	fact[0] = modint(1);
	for (int i = 0; i < max_n - 1; i++) {
		fact[i + 1] = fact[i] * modint(i + 1);
	}
	factinv[max_n - 1] = modint(1) / fact[max_n - 1];
	for (int i = max_n - 2; i >= 0; i--) {
		factinv[i] = factinv[i + 1] * modint(i + 1);
	}
}
modint comb(int a, int b) {
	if (a < 0 || b < 0 || a < b)return 0;
	return fact[a] * factinv[b] * factinv[a - b];
}
using mP = pair<modint, modint>;

ll t3[31];
void solve() {
	t3[0] = 1;
	rep(i, 30) {
		t3[i + 1] = t3[i] * 3;
	}
		ll x[2], y[2];
		cin >> x[0] >> y[0] >> x[1] >> y[1];
		rep(i, 2) {
			x[i]--; y[i]--;
		}

		if (x[0] > x[1]) {
			swap(x[0], x[1]);
			swap(y[0], y[1]);
		}
		ll ans = abs(x[1] - x[0]) + abs(y[1] - y[0]);
		for (int t = 29; t >= 0; t--) {
			ll dx = x[0] / t3[t+1];
			ll gdx = x[1] / t3[t + 1];
			ll dy = y[0] / t3[t + 1];
			ll gdy = y[1] / t3[t + 1];
			
			if (dx == gdx) {
				ll rx = x[0] % t3[t + 1];
				ll grx = x[0] % t3[t + 1];
				if (t3[t] <= rx && rx < 2 * t3[t] && t3[t] <= grx && grx < 2 * t3[t]) {
					bool prob = true;
					if (dy == gdy) {
						ll ry = y[0] % t3[t + 1];
						ll gry = y[1] % t3[t + 1];
						prob = false;
						if (ry < t3[t] && gry >= 2 * t3[t])prob = true;
						if (gry < t3[t] && ry >= 2 * t3[t])prob = true;
					}
					else {
						ll yy = abs(y[1] - y[0]);
						if (yy < 2 * t3[t])prob = false;
					}
					if (prob) {
						ll ori = abs(x[1] - x[0]);
						ll nexu = rx - t3[t] + 1 + grx - t3[t] + 1;
						ll nexd = 2 * t3[t] - rx + 2 * t3[t] - grx;
						ll ad = min(nexu, nexd) - ori;
						cout << ans + ad << endl; return;
					}
				}
			}
			if (dy == gdy) {
				ll ry = y[0] % t3[t + 1];
				ll gry = y[1] % t3[t + 1];
				if (t3[t] <= ry && ry < 2 * t3[t] && t3[t] <= gry && gry < 2 * t3[t]) {
					bool prob = true;
					if (dx == gdx) {
						ll rx = x[0] % t3[t + 1];
						ll grx = x[1] % t3[t + 1];
						prob = false;
						if (rx < t3[t] && grx >= 2 * t3[t])prob = true;
						if (grx < t3[t] && rx >= 2 * t3[t])prob = true;
					}
					else {
						ll xx = abs(x[1] - x[0]);
						if (xx < 2 * t3[t])prob = false;
					}
					if (prob) {
						ll ori = abs(y[1] - y[0]);
						ll nexu = ry - t3[t] + 1 + gry - t3[t] + 1;
						ll nexd = 2 * t3[t] - ry + 2 * t3[t] - gry;
						ll ad = min(nexu, nexd) - ori;
						cout << ans + ad << endl; return;
					}
				}
			}
		}
		cout << ans << endl;
}


signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	//cout << fixed << setprecision(7);
	//init_f();
	//init();
	int t; cin >> t; rep(i, t)solve();
	//solve();
	stop
		return 0;
}