Problem: Find the diameter of a convex polygon g. In other words, find a pair of points that have maximum distance between them.

Constraints

* 3 ≤ n ≤ 80000
* -100 ≤ xi, yi ≤ 100
* No point in the g will occur more than once.

Input


n
x1 y1
x2 y2
:
xn yn


The first integer n is the number of points in g.

In the following lines, the coordinate of the i-th point pi is given by two real numbers xi and yi. The coordinates of points are given in the order of counter-clockwise visit of them. Each value is a real number with at most 6 digits after the decimal point.

Output

Print the diameter of g in a line. The output values should be in a decimal fraction with an error less than 0.000001.

Examples

Input

3
0.0 0.0
4.0 0.0
2.0 2.0


Output

4.00


Input

4
0.0 0.0
1.0 0.0
1.0 1.0
0.0 1.0


Output

1.414213562373

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <deque>
#include <queue>
#include <array>
#include <set>
#include <map>
#include <cmath>
#include <complex>
#include <algorithm>
#include <numeric>
#include <utility>
#include <tuple>
#include <bitset>
#include <cstdint>
#include <cassert>
#include <random>
#include <iterator>

using namespace std;
using i64 = int64_t;
using i32 = int32_t;
template<class T, class U> void init_n(vector<T>& v, size_t n, U x) 
{ v = vector<T>(n, x); }
template<class T> void init_n(vector<T>& v, size_t n) { init_n(v, n, T()); }
template<class T> void read_n(vector<T>& v, size_t n, size_t o = 0) 
{ v = vector<T>(n+o); for (size_t i=o; i<n+o; ++i) cin >> v[i]; }
template<class T> void read_n(T a[], size_t n, size_t o = 0)
{ for (size_t i=o; i<n+o; ++i) cin >> a[i]; }
// template<class T> T gabs(const T& x) { return max(x, -x); }
// #define abs gabs

int main() {
    int n;
    cin >> n;
    vector<double> x(n), y(n);
    for (int i = 0; i < n; ++i) {
        cin >> x[i] >> y[i];
    }
    double ans = 0;
    for (int i = 0, j = 0; i < n; ++i) {
        while (hypot(x[j] - x[i], y[j] - y[i]) 
            < hypot(x[(j + 1) % n] - x[i], y[(j + 1) % n] - y[i])) {
            j = (j + 1) % n;
        }
        ans = max(ans, hypot(x[j] - x[i], y[j] - y[i]));
    }
    printf("%.10f\n", ans);
    return 0;
}