//方法1 二分

//方法1 二分
int get(int a, int b) {
    int l = 1, r = 1e9 + 1; //这里取到1e9 + 1 是因为下面可能有b - 1,当b取1时
    //为0的情况,所以当r = 1e9 + 1可以让成立这个情况
    while (l < r) {
        int mid = l + r >> 1;
        if ((a / mid) <= b) r = mid;
        else l = mid + 1;
    }
    return l;
}

int main() {
    scanf("%d", &n);
    int min_v = 1, max_v = 1e9;
    while (n--) {
        int a, b;
        scanf("%d%d", &a, &b);
        min_v = max(min_v, get(a, b));
        max_v = min(max_v, get(a, b - 1) - 1);
    }
    printf("%d %d", min_v, max_v);
    return 0;
}

//方法1 推公式

#include<bits/stdc++.h>
#include<set>
using namespace std;

int n;

int get_max(int a, int b) {
    return a / b;
}

int get_min(int a, int b) {
    return a / (b + 1) + 1;
}

int main() {
    scanf("%d", &n);
    int min_v = 1, max_v = 1e9;
    while (n--) {
        int a, b;
        scanf("%d%d", &a, &b);
        min_v = max(min_v, get_min(a, b));
        max_v = min(max_v, get_max(a, b));
    }
    printf("%d %d", min_v, max_v);
    return 0;
}