#include<iostream>
#include<algorithm>
using namespace std;
const int N = 1e5 + 5; 
int n,k;
int ranc[N][2];
bool check(int mid) {
    int res = 0;
    for (int i = 0;i < n;i++) {
    //防止长方形的长或宽小于正方形的边长，因为题中的巧克力是不能拼凑的
        res += (ranc[i][0] / mid) * (ranc[i][1] / mid);
    }
    return res >= k;
}
int main() {
    cin >> n >> k;
    for (int i = 0;i < n;i++) {
        cin >> ranc[i][0] >> ranc[i][1];
    }
    int l = 1, r = 1e5;
    while (l < r) {
        int mid = (l + r + 1) / 2;//写法为向上取整
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    cout << l << endl;
}