#include<bits/stdc++.h>

using namespace std;

const int N = 1010;
int n;
struct Person {
    int left, right;
} person[N];

bool cmp(const Person &a, const Person &b) {
    return a.left * a.right < b.left * b.right;
}

//高精度乘法
vector<int> mul(vector<int> &A, int b) {
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size() || t; i++) {
        if (i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

//高精度除法
vector<int> div(vector<int> &A, int b, int &r) {
    vector<int> C;
    r = 0;
    for (int i = A.size() - 1; i >= 0; i--) {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    reverse(C.begin(), C.end());
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

//获取两个vector<int>中较大的那个
vector<int> max_vec(vector<int> a, vector<int> b) {
    if (a.size() > b.size()) return a;
    if (a.size() < b.size()) return b;
    for (int i = a.size() - 1; i >= 0; i--) {
        if (a[i] > b[i]) return a;
        if (a[i] < b[i]) return b;
    }
    return a;
}

int main() {
    //优化输入
    ios::sync_with_stdio(false);
    cin >> n;
    //输入
    for (int i = 0; i <= n; i++)
        cin >> person[i].left >> person[i].right;

    //排序,国王不参加排序
    sort(person + 1, person + n + 1, cmp);

    //连乘各放入国王
    vector<int> lcj;
    lcj.push_back(person[0].left);

    //结果
    vector<int> res;
    res.push_back(0);//在做高精度时，也需要进行一步初始化

    for (int i = 1; i <= n; i++) {
        int r;//余数，本题中因为只需要下取整，所以余数无意义
        res = max_vec(res, div(lcj, person[i].right, r));
        lcj = mul(lcj, person[i].left);
    }
    //倒序输出高精度数组中的值
    for (int i = res.size() - 1; i >= 0; i--)
        printf("%d", res[i]);
    return 0;
}