Problem:
Score: 600 points

Problem Statement

Takahashi and Aoki are training for long-distance races in an infinitely long straight course running from west to east.

They start simultaneously at the same point and moves as follows towards the east:

Takahashi runs A_1 meters per minute for the first T_1 minutes, then runs at A_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever.

Aoki runs B_1 meters per minute for the first T_1 minutes, then runs at B_2 meters per minute for the subsequent T_2 minutes, and alternates between these two modes forever.

How many times will Takahashi and Aoki meet each other, that is, come to the same point? We do not count the start of the run. If they meet infinitely many times, report that fact.

Constraints

1 \leq T_i \leq 100000

1 \leq A_i \leq 10^{10}

1 \leq B_i \leq 10^{10}

A_1 \neq B_1

A_2 \neq B_2

All values in input are integers.

Input

Input is given from Standard Input in the following format:

T_1 T_2
A_1 A_2
B_1 B_2

Output

Print the number of times Takahashi and Aoki will meet each other.

If they meet infinitely many times, print infinity instead.

Sample Input 1

1 2
10 10
12 4

Sample Output 1

1

They will meet just once, \frac{4}{3} minutes after they start, at \frac{40}{3} meters from where they start.

Sample Input 2

100 1
101 101
102 1

Sample Output 2

infinity

They will meet 101, 202, 303, 404, 505, 606, ... minutes after they start, that is, they will meet infinitely many times.

Sample Input 3

12000 15700
3390000000 3810000000
5550000000 2130000000

Sample Output 3

113

The values in input may not fit into a 32-bit integer type.

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

#include <bits/stdc++.h>

using ll = long long;
#define int long long

namespace ext {
	template <class T>
	struct vec_type;
	template <class T>
	constexpr bool is_vec_v;
	using namespace std;
	template<class T>
	struct Vector : std::vector<T> {
		using std::vector<T>::vector;
		char dil = ' ';
		Vector<T> accum() {
			Vector<T> res(this->size() + 1);
			for (size_t i = 0; i < this->size(); i++) {
				res[i+1] = (*this)[i] + res[i];
			}
			return res;
		}
		template<class U = T>
		typename enable_if<is_vec_v<U>, Vector<T>>::type accum2D() {
			const size_t H = this->size();
			const size_t W = (*this)[0].size();
			using S = typename vec_type<U>::type;
			Vector<T> res(H + 1, Vector<S>(W + 1));
			for (size_t i = 0; i < H; i++) {
				for (size_t j = 0; j < W; j++) {
					res[i+1][j+1] = (*this)[i][j];
					res[i+1][j+1] += res[i][j+1];
					res[i+1][j+1] += res[i+1][j];
					res[i+1][j+1] -= res[i][j];
				}
			}
			return res;
		}
		Vector<T> zaatsu() {
			map<T, int> m;
			for (auto& e: *this) m[e] = 1;
			int count = 0;
			for (auto& e: m) e.second = count++;
			Vector<T> res = *this;
			for (auto& e: res) e = m[e];
			return res;
		}
		friend istream& operator>>(istream& is, Vector<T>& t) {
			for (auto& e: t) cin >> e;
			return is;
		}
		friend ostream& operator<<(ostream& os, Vector<T>& t) {
			t.print(' ', os);
			return os;
		}
		void print(char del = ' ', ostream& os = cout) {
			for (size_t i = 0; i < this->size(); i++) {
				os << (*this)[i];
				if (i + 1 != this->size()) os << del;
				os.flush();
			}
		}
		void println(char del = ' ', ostream& os = cout) {
			print(del, os);
			cout << endl;
		}
		long long crossing_merge() {
			auto t = *this;
			return t.crossing_merge_sub();
		}
		// [min_v, max_v)
		long long crossing_bit(T min_v = -1, T max_v = -1) {
			if (min_v == -1) min_v = this->min();
			if (max_v == -1) max_v = this->max() + 1;
			Vector<T> bit(max_v - min_v + 1, 0);
			long long res = 0;
			for (auto e: *this) {
				e = max_v - e;
				for (size_t x = e - 1; x; x -= x & -x) {
					res += bit[x];
				}
				for (size_t x = e; x < bit.size() ; x += x & -x) {
					bit[x]++;
				}
			}
			return res;
		}
		bool permutation() {
			return next_permutation(this->begin(), this->end());
		}
		T& max() {
			return *max_element(this->begin(), this->end());
		}
		T& min() {
			return *min_element(this->begin(), this->end());
		}
private:
		long long crossing_merge_sub() {
			if (this->size() == 1) return 0;
			int x = this->size() / 2;
			Vector<T> left(this->begin(), this->begin() + x);
			Vector<T> right(this->begin() + x, this->end());
			long long res = left.crossing_merge_sub() + right.crossing_merge_sub();
			left.push_back(numeric_limits<T>::max()); right.push_back(numeric_limits<T>::max());
			size_t l = 0, r = 0;
			while (l + r < this->size()) {
				if (left[l] < right[r]) {
					(*this)[l+r] = left[l];
					l++;
				} else {
					(*this)[l+r] = right[r];
					r++;
					res += x - l;
				}
			}
			return res;
		}
	};
	template <class T>
	struct vec_type : std::false_type {};
	template <class T>
	struct vec_type<Vector<T>> : std::true_type {
		using type = T;
	};
	template <class T>
	constexpr bool is_vec_v = vec_type<T>::value;
}

// [a -> b-1]
#define reps(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
// [0 -> a-1]
#define rep(i, a) reps(i, 0, (a))
// [a-1 -> b]
#define rreps(i, a, b) for (int i = (int)((a)-1); i >= (int)(b); i--)
// [a-1 -> 0]
#define rrep(i, a) rreps(i, a, 0)
#define all(v) (v).begin(), (v).end()
// next_permutation(all(v))
#define PERM(v) next_permutation(all(v))
/*sort(all(v));
 * (v).erase(unique(all(v)), v.end())*/
#define UNIQUE(v)\
	sort(all(v));\
	(v).erase(unique(all(v)), v.end())
#define MINV(v) min_element(all(v))
#define MAXV(v) max_element(all(v))
#define MIN3(a, b, c) min(min(a, b), c)
#define MIN4(a, b, c, d) min(MIN3(a, b, c), d)
#define MIN5(a, b, c, d, e) min(MIN4(a, b, c, d), e)
#define MIN6(a, b, c, d, e, f) min(MIN5(a, b, c, d, e), f)
#define MAX3(a, b, c) max(max(a, b), c)
#define MAX4(a, b, c, d) max(MAX3(a, b, c), d)
#define MAX5(a, b, c, d, e) max(MAX4(a, b, c, d), e)
#define MAX6(a, b, c, d, e, f) max(MAX5(a, b, c, d, e), f)
// b is [a, c)
#define RANGE(a, b, c) ((a) <= (b) && (b) < (c))
// c is [a, e) && d is [b, f)
#define RANGE2D(a, b, c, d, e, f) (RANGE((a), (c), (e)) && RANGE((b), (d), (f)))
#define chmin(a, b) a = min(a, (b))
#define chmin3(a, b, c) a = MIN3(a, (b), (c))
#define chmin4(a, b, c, d) a = MIN4(a, (b), (c), (d))
#define chmin5(a, b, c, d, e) a = MIN5(a, (b), (c), (d), (e))
#define chmin6(a, b, c, d, e, f) a = MIN6(a, (b), (c), (d), (e), (f))
#define chmax(a, b) a = max(a, (b))
#define chmax3(a, b, c) a = MAX3(a, (b), (c))
#define chmax4(a, b, c, d) a = MAX4(a, (b), (c), (d))
#define chmax5(a, b, c, d, e) a = MAX5(a, (b), (c), (d), (e))
#define chmax6(a, b, c, d, e, f) a = MAX6(a, (b), (c), (d), (e), (f))
#define fcout cout << fixed << setprecision(15)
#define YES(f) cout << ((f) ? YES_STR : NO_STR) << endl;

using namespace std;
using namespace ext;

int T1, T2, A1, A2, B1, B2;
int X;

int isok(int x) {
	if (x == 0) return A1 * T1 - (B1 * T1 + X * x);
	return -((B1 * T1 - A1 * T1) / x + X);
}

#define cin input
#define cout output

// meguru<T>(ok, ng, isok)で呼ぶ
template<class T>
T meguru(T ok, T ng) {
	while (abs(ok - ng) > 1) {
		T mid = (ok + ng) >> 1;
		(isok(mid) >= 0 ? ok : ng) = mid;
	}
	return ok;
}

void solve(istream& input, ostream& output) {
	cin >> T1 >> T2;
	cin >> A1 >> A2;
	cin >> B1 >> B2;
	if (T1 * A1 + T2 * A2 == T1 * B1 + T2 * B2) {
		cout << "infinity" << endl;
		return;
	}
	if (T1 * A1 + T2 * A2 > T1 * B1 + T2 * B2) {
		swap(A1, B1);
		swap(A2, B2);
	}
	X = -(T1 * A1 + T2 * A2 - T1 * B1 - T2 * B2);
	int ng = 1;
	while (isok(ng) >= 0) ng *= 2;
	int r = meguru(-1LL, ng);
	int res = max(0LL, r * 2 + 1);
	if (r > -1 && isok(r) == 0) res--;
	cout << res << endl;
}

#undef cin
#undef cout

#ifndef CHECK_CPP

signed main() {
	solve(cin, cout);
}

#endif