#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
# define rep(i,a,b) for(int i=(a); i<=(b); ++i)
# define drep(i,a,b) for(int i=(a); i>=(b); --i)
typedef long long int_;
inline int readint(){
	int a = 0; char c = getchar(), f = 1;
	for(; c<'0'||c>'9'; c=getchar())
		if(c == '-') f = -f;
	for(; '0'<=c&&c<='9'; c=getchar())
		a = (a<<3)+(a<<1)+(c^48);
	return a*f;
}

const int MaxN = 400005;
pair<int,int> a[MaxN];
pair<int,int> b[MaxN];
vector<int> v;

int main(){
	int n = readint();
	rep(i,1,n<<1){
		a[i].first = readint();
		a[i].second = i;
		b[i] = a[i]; // copy
	}
	sort(b+1,b+(n<<1)+1);
	rep(i,1,n<<1)
		if(v.empty())
			v.push_back(i), putchar('(');
		else{
			int t = v.back();
			if((a[t] > b[n]) == (a[i] > b[n]))
				v.push_back(i), putchar('(');
			else v.pop_back(), putchar(')');
		}
	putchar('\n');
	return 0;
}

#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);
	}
}
