Problem:
Score : 300 points

Problem Statement

AtCoDeer the deer is seeing a quick report of election results on TV.
Two candidates are standing for the election: Takahashi and Aoki.
The report shows the ratio of the current numbers of votes the two candidates have obtained, but not the actual numbers of votes.
AtCoDeer has checked the report N times, and when he checked it for the i-th (1≦i≦N) time, the ratio was T_i:A_i.
It is known that each candidate had at least one vote when he checked the report for the first time.

Find the minimum possible total number of votes obtained by the two candidates when he checked the report for the N-th time.
It can be assumed that the number of votes obtained by each candidate never decreases.

Constraints

1≦N≦1000

1≦T_i,A_i≦1000 (1≦i≦N)

T_i and A_i (1≦i≦N) are coprime.

It is guaranteed that the correct answer is at most 10^{18}.

Input

The input is given from Standard Input in the following format:

N
T_1 A_1
T_2 A_2
:
T_N A_N

Output

Print the minimum possible total number of votes obtained by Takahashi and Aoki when AtCoDeer checked the report for the N-th time.

Sample Input 1

3
2 3
1 1
3 2

Sample Output 1

10

When the numbers of votes obtained by the two candidates change as 2,3 → 3,3 → 6,4, the total number of votes at the end is 10, which is the minimum possible number.

Sample Input 2

4
1 1
1 1
1 5
1 100

Sample Output 2

101

It is possible that neither candidate obtained a vote between the moment when he checked the report, and the moment when he checked it for the next time.

Sample Input 3

5
3 10
48 17
31 199
231 23
3 2

Sample Output 3

6930

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

#include "bits/stdc++.h"
using namespace std;

typedef long long ll;
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)

//* 便利な変数
namespace {
	int dx4[] = { 1, -1, 0, 0 };
	int dy4[] = { 0, 0, 1, -1 };

	int dx8[] = { 1, -1, 0, 0, 1, 1, -1, -1 };
	int dy8[] = { 0, 0, -1, 1, -1, 1, -1, 1 };

	int mDays[] = { 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

	ll A, B, C, D, E, F, G, H, I, J, K, L, M,
		N, O, P, Q, R, S, T, U, V, W, X, Y, Z;
}
template <typename T>
vector<T> INP(ll n)
{
	vector<T> x;
	REP(i, n) {
		T tmp; cin >> tmp;
		x.push_back(tmp);
	}
	return move(x);
}
//* n文字1行の文字列を入力，一文字ごとの配列を返す
vector<char> SPRIT_STRING(ll n)
{
	string str; cin >> str;
	vector<char> cs(n);
	REP(i, n) cs[i] = str[i];
	return move(cs);
}
//* 文字列中から文字列を検索して別の文字列に置換する
void strReplace(std::string& str, const std::string& from, const std::string& to) {
	std::string::size_type pos = 0;
	while (pos = str.find(from, pos), pos != std::string::npos) {
		str.replace(pos, from.length(), to);
		pos += to.length();
	}
}
//* 素数判定 is_prime<unsigned>(N)
template<typename T, std::enable_if_t<std::is_unsigned<T>::value, std::nullptr_t> = nullptr>
bool is_prime(const T n) {
	if (n < 4) return n == 2 || n == 3;
	if (n % 2 == 0 || n % 3 == 0 || (n % 6 != 1 && n % 6 != 5)) return false;
	for (T i = 5; i * i <= n; i += 6) if (n % i == 0 || n % (i + 2) == 0) return false;
	return true;
}
//* 組み合わせ計算
inline unsigned long long NChooseK(const unsigned long long& n, const unsigned long long& k)
{
	if (n  < k) return 0;
	if (0 == n) return 0;
	if (0 == k) return 1;
	if (n == k) return 1;
	if (1 == k) return n;
	typedef unsigned long long value_type;
	value_type* table = new value_type[static_cast<std::size_t>(n * n)];
	std::fill_n(table, n * n, 0);
	class n_choose_k_impl
	{
	public:

		n_choose_k_impl(value_type* table, const value_type& dimension)
			: table_(table),
			dimension_(dimension)
		{}

		inline value_type& lookup(const value_type& n, const value_type& k)
		{
			return table_[dimension_ * n + k];
		}

		inline value_type compute(const value_type& n, const value_type& k)
		{
			if ((0 == k) || (k == n))
				return 1;
			value_type v1 = lookup(n - 1, k - 1);
			if (0 == v1)
				v1 = lookup(n - 1, k - 1) = compute(n - 1, k - 1);
			value_type v2 = lookup(n - 1, k);
			if (0 == v2)
				v2 = lookup(n - 1, k) = compute(n - 1, k);
			return v1 + v2;
		}

		value_type* table_;
		value_type dimension_;
	};
	value_type result = n_choose_k_impl(table, n).compute(n, k);
	delete[] table;
	return result;
}
//* 座標nx, nyがWidth,Heightの領域内にあるかどうかのチェック
inline bool rangeCheck2D(int nx, int ny, int Width, int Height)
{
	return nx >= 0 and nx < Width and ny >= 0 and ny < Height;
}
//* bit全探索
/*
for (int i = 0; i < 1<<N; i++) {
	REP(j, N)
		if ((1 & i >> j) == 1) {;}
}
*/

namespace arc062a {
	void arc062a()
	{
		cin >> N;
		ll ct, ca;
		REP(i, N) {
			ll t, a; cin >> t >> a;
			if (i == 0 or (t >= ct && a >= ca)) {
				ct = t; ca = a;
			}
			else {
				ll mult = 1, mula = 1;
				if (t < ct)
					mult = ct / t + ((ct%t == 0) ? 0 : 1);
				if (a < ca)
					mula = ca / a + ((ca%a == 0) ? 0 : 1);
				ll mulv = max(mult, mula);
				ct = t * mulv; ca = a * mulv;
			}
		}
		cout << ct + ca << endl;
	}
}

namespace arc06 {

}

int main()
{
	//abc079b::abc079b();
	//abc081d::abc081d();
	arc062a::arc062a();
	return 0;
}