#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <algorithm>
#include <iomanip>
#include <math.h>
#include <string.h>
#include <cstdio>
#include <tuple>
#include <numeric>
#include <functional>
using namespace std; using ll = long long; using ld = long double;  using pll = pair<ll, ll>;
using vl = vector<ll>; using vll = vector<vl>; using vlll = vector<vll>; using vpll = vector<pll>;
using vs = vector<string>; using tll = tuple<ll, ll, ll>; using vtll = vector<tll>;
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
const ld PI = 3.1415926535897932;
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define co(i) cout << i << endl;
void in() {}
void debug_out() { cerr << endl; }
template<typename Head, typename... Tail>
void debug_out(Head h, Tail... t) {
	cerr << " " << h;
	if (sizeof...(t) > 0) cerr << " :";
	debug_out(t...);
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> vec) {
	for (size_t i = 0; i < vec.size(); i++)os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	return os;
}
ll ctoi(char c) {
	switch (c) {
	case '0': return 0; case '1': return 1; case '2': return 2;
	case '3': return 3; case '4': return 4; case '5': return 5;
	case '6': return 6; case '7': return 7; case '8': return 8;
	case '9': return 9;  default: return 0;
	}
}
bool pairCompare(const pll firstElof, pll secondElof)
{
	return firstElof.second < secondElof.second;
}
//AtCoder以外ではコメントアウトを忘れずに！
//#include <atcoder/all>
//using namespace atcoder;
//**Snippetリスト**//
//rep, vin, all, iteLoop, bitSearch, bitList, nod, LIS, digitDP, treeDP, Rerooting//
//gcdlcm, isPrime, eratos, primeFactorize, Npow, combination, divisor, modinv, doubling//
//dfs, bfs, dijkstra, WarshallFloyd, BellmanFord, UnionFind, Kruskal, RMQ, LCA, FordFulkerson//
//**AC Libraryリスト**//
//pow_mod, inv_mod, floor_sum, convolution, convolution_ll, modint//
//fenwicktree, segtree, lazysegtree, string//
ll i, j, k, l; ll N, M, K, H, W, X, Y, Z, R, Q;
ll MOD = 1000000007, INF = 1LL << 60, ans = 0, z = 0, o = 1;
vector<vtll> path;
//***********//
inline long long mod2(long long a, long long m) {
	return (a % m + m) % m;
}

// 拡張 Euclid の互除法
// ap + bq = gcd(a, b) となる (p, q) を求め、d = gcd(a, b) をリターンします
long long extGcd(long long a, long long b, long long& p, long long& q) {
	if (b == 0) { p = 1; q = 0; return a; }
	long long d = extGcd(b, a % b, q, p);
	q -= a / b * p;
	return d;
}

// 中国剰余定理
// リターン値を (r, m) とすると解は x ≡ r (mod. m)
// 解なしの場合は (0, -1) をリターン
pair<long long, long long> ChineseRem(long long b1, long long m1, long long b2, long long m2) {
	long long p, q;
	long long d = extGcd(m1, m2, p, q); // p is inv of m1/d (mod. m2/d)
	if ((b2 - b1) % d != 0) return make_pair(0, -1);
	long long m = m1 * (m2 / d); // lcm of (m1, m2)
	long long tmp = (b2 - b1) / d * p % (m2 / d);
	long long r = mod2(b1 + m1 * tmp, m);
	return make_pair(r, m);
}
int main() {
	ll T;
	cin >> T;
	for (l = 0; l < T; l++) {
		ll X, Y, P, Q;
		cin >> X >> Y >> P >> Q;
		ll ans = INF;
		for (i = 0; i < Y; i++) {
			for (j = 0; j < Q; j++) {
				ll m1 = X * 2 + Y * 2;
				ll b1 = X + i;
				ll m2 = P + Q;
				ll b2 = P + j;
				pll Ans = ChineseRem(b1, m1, b2, m2);
				//debug(b1,m1, b2,m2, Ans.first,Ans.second);
				if (Ans.second != -1) {
					chmin(ans, Ans.first);
				}
			}
		}
		//debug(1000000000999999999 - ans);
		if (ans == INF) cout << "infinity" << endl;
		else cout << ans << endl;
		//debug(ChineseRem(3, 2, 5, 3).first);
	}
}

#include <bits/stdc++.h>
#define DEBUG if(0)
#define lli __int128
#define ldouble long double
using namespace std;

/*
(2x + 2y)n + x <= t < (2x + 2y)n + x + y
  range = y (max 500)
(p + q)n + p <= t < (p + q)(n + 1)
  range = q (max 500)

try all combinations of mod (500^2)

t - (2x + 2y)n - x == remY
t - (2x + 2y)n = remY + x
t = remY + x + (2x + 2y)n1

t - (p + q)n - p == remQ
t - (p + q)n = remQ + p
t = remQ + p + (p + q)n2

remQ + p + (p + q)n2 = remY + x + (2x + 2y)n1
(p + q)n2 - (2x + 2y)n1 = remY + x - remQ - p

*/

lli read() {
    lli x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
        if (ch == '-') f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
void print(lli x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
}

lli abslol(lli num)
{
  return num < 0 ? -num : num;
}

lli gcd(lli a, lli b, lli& x, lli& y)
{
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  lli x1, y1;
  lli d = gcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}

bool find_any_solution(lli a, lli b, lli c, lli &x0, lli &y0, lli &g)
{
  g = gcd(abslol(a), abslol(b), x0, y0);
  if (c % g) {
    return false;
  }

  x0 *= c / g;
  y0 *= c / g;
  if (a < 0) x0 = -x0;
  if (b < 0) y0 = -y0;
  return true;
}

const lli inf = LONG_LONG_MAX;
lli x, y, p, q;


lli findMinSolution(lli remY, lli remQ)
{
  // t - (2x + 2y)n - x == remY
  // t - (2x + 2y)n = remY + x
  // t = remY + x + (2x + 2y)x0

  // t - (p + q)n - p == remQ
  // t - (p + q)n = remQ + p
  // t = remQ + p + (p + q)y0

  // remQ + p + (p + q)y0 = remY + x + (2x + 2y)x0
  // -(2x + 2y)x0 + (p + q)y0 = remY + x - remQ - p
  lli x0, y0, g;
  lli a = -(2*x + 2*y), b = p + q, c = remY + x - remQ - p;
  if (find_any_solution(a, b, c, x0, y0, g))
  {
    lli t1 = remY + x + (2*x + 2*y)*x0, t2 = remQ + p + (p + q)*y0;
    bool isNegative = a / g < 0;

    // lli lo = 0, hi = 1e18;
    // while (lo < hi)
    // {
    //   lli mid = (lo + hi + 1) >> 1;
    //   if ((y0 + mid*(a / g)) != 0 && LONG_LONG_MAX / abslol(y0 + mid*(a / g)) < (p + q))
    //     hi = mid - 1;
    //   else
    //     lo = mid;
    // }
    // lli maxHi = lo;

    // lo = -1e18, hi = 0;
    // while (lo < hi)
    // {
    //   lli mid = (lo + hi) >> 1;
    //   if ((y0 + mid*(a / g)) != 0 && LONG_LONG_MAX / abslol(y0 + mid*(a / g)) < (p + q))
    //     lo = mid + 1;
    //   else
    //     hi = mid;
    // }
    // lli minLo = lo;

    lli lo = -1e18, hi = 1e18;
    while (lo < hi)
    {
      lli mid = (lo + hi + isNegative) >> 1;
      lli e = remQ + p + (p + q)*(y0 + mid*(a / g));

      // DEBUG printf("\t\t%lld %lld\n", mid, e);
      if (isNegative)
      {
        if (e >= 0) lo = mid;
        else hi = mid - 1;
      }
      else
      {
        if (e >= 0) hi = mid;
        else lo = mid + 1;
      }
    }

    x0 -= lo * b / g;
    y0 += lo * a / g;

    // DEBUG printf("\t%lld %lld - %lld %lld | %lld %lld\n", remY, remQ, t1, t2, 0LL, b / g);
    return remQ + p + (p + q)*y0;
  }
  else
    return inf;
}

int main()
{
  int t; scanf("%d", &t);
  for (int tt = 1; tt <= t; tt++)
  {
    // scanf("%lld %lld %lld %lld", &x, &y, &p, &q);
    x = read();
    y = read();
    p = read();
    q = read();
    lli ans = inf;
    for (int i = 0; i < y; i++)
      for (int j = 0; j < q; j++)
        ans = min(ans, findMinSolution(i, j));

    if (ans == inf) printf("infinity\n");
    else
    {
      print(ans);
      printf("\n");
    }
  }

  return 0;
}