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 <bits/stdc++.h>
//#define int long long

using namespace std;
using LL = long long;
using P = pair<int, int>;
using Tapris = tuple<int, int, int>;

#define FOR(i, a, n) for(int i = (int)(a); i < (int)(n); ++i)
#define REP(i, n) FOR(i, 0, n)

#define pb(a) push_back(a)
#define all(x) (x).begin(),(x).end()

const int INF = (int)1e9;
const LL INFL = (LL)1e15;
const int MOD = 1e9 + 7;

int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};

typedef long double LD;
typedef complex<LD> Point;
typedef pair<Point, Point> Line;
typedef vector<Point> Polygon;
const LD EPS = 1e-10;
#define X real() // x座標を取得
#define Y imag() // y座標を取得
#define LE(n,m) ((n) < (m) + EPS)
#define GE(n,m) ((n) + EPS > (m))
#define EQ(n,m) (abs((n)-(m)) < EPS)

// 内積 Dot(a, b) = |a||b|cosθ
LD Dot(Point a, Point b){
    return (conj(a)*b).X;
}

// 外積 Cross(a, b) = |a||b|sinθ
LD Cross(Point a, Point b){
    return (conj(a)*b).Y;
}

int CCW(Point a, Point b, Point c){
    b -= a; c -= a;
    if (Cross(b, c) > 0)   return +1;       // counter clockwise
    if (Cross(b, c) < 0)   return -1;       // clockwise
    if (Dot(b, c) < 0)     return +2;       // c--a--b on line
    if (norm(b) < norm(c)) return -2;       // a--b--c on line
    return 0;
}

//*********************************************
//          点と線(Point and Line)            *
//*********************************************

// 交差判定 (Isec) ****************************
// 点　 := 平面座標にある点
// 直線 := 点と点を通るどこまでも続く線
// 線分 := 点と点を結んでその両端で止まっている線

// 直線と点
bool IsecLP(Point a1, Point a2, Point b){
    return abs(CCW(a1, a2, b)) != 1;
}

// 直線と直線
bool IsecLL(Point a1, Point a2, Point b1, Point b2) {
    return !IsecLP(a2-a1, b2-b1, 0) || IsecLP(a1, b1, b2);
}

// 直線と線分
bool IsecLS(Point a1, Point a2, Point b1, Point b2) {
    return Cross(a2-a1, b1-a1) * Cross(a2-a1, b2-a1) < EPS;
}

// 線分と線分
bool IsecSS(Point a1, Point a2, Point b1, Point b2) {
    return CCW(a1, a2, b1)*CCW(a1, a2, b2) <= 0 && CCW(b1, b2, a1)*CCW(b1, b2, a2) <= 0;
}

// 線分と点
bool IsecSP(Point a1, Point a2, Point b) {
    return !CCW(a1, a2, b);
}

// ********************************************


// 距離 (Dist) ********************************
// 点pの直線aへの射影点を返す
Point Proj(Point a1, Point a2, Point p){
    return a1 + Dot(a2-a1, p-a1) / norm(a2-a1) * (a2-a1);
}

// 点pの直線aへの反射点を返す
Point Reflection(Point a1, Point a2, Point p){
    return 2.0L*Proj(a1, a2, p) - p;
}

// 点と点
LD DistPP(Point a, Point b){
    return abs(a-b);
}

// 直線と点
LD DistLP(Point a1, Point a2, Point p){
    return abs(Proj(a1, a2, p) - p);
}

// 直線と直線
LD DistLL(Point a1, Point a2, Point b1, Point b2) {
    return IsecLL(a1, a2, b1, b2) ? 0 : DistLP(a1, a2, b1);
}

// 直線と線分
LD DistLS(Point a1, Point a2, Point b1, Point b2) {
    return IsecLS(a1, a2, b1, b2) ? 0 : min(DistLP(a1, a2, b1), DistLP(a1, a2, b2));
}

// 線分と点
LD DistSP(Point a1, Point a2, Point p) {
    if(Dot(a2-a1,p-a1) < EPS) return abs(p-a1);
    if(Dot(a1-a2,p-a2) < EPS) return abs(p-a2);
    return abs(Cross(a2-a1,p-a1)) / abs(a2-a1);
}

// 線分と線分
LD DistSS(Point a1, Point a2, Point b1, Point b2) {
    if(IsecSS(a1, a2, b1, b2)) return 0;
    return min(min(DistSP(a1, a2, b1), DistSP(a1, a2, b2)), min(DistSP(b1, b2, a1), DistSP(b1, b2, a2)));
}

// ********************************************


// 2直線の交点 (CrossPoint) *******************
Point CrossPointLL(Point a1, Point a2, Point b1, Point b2){
    LD d1 = Cross(b2-b1, b1-a1);
    LD d2 = Cross(b2-b1, a2-a1);
    if (EQ(d1, 0) && EQ(d2, 0)) return a1;
    if (EQ(d2, 0)) throw "not exist crosspoint";
    return a1 + d1/d2 * (a2-a1);
}
// ********************************************


//*********************************************
//             多角形(Polygon)                *
//*********************************************

namespace std {
    bool operator < (Point a, Point b){
        return a.X != b.X ? a.X < b.X : a.Y < b.Y;
    }
}

// 多角形の面積 (PolygonErea) *****************
LD PolygonErea(Polygon p){
    LD area = 0;
    int n = p.size();
    for(int i = 0; i < n; i++){
        area += Cross(p[i], p[(i+1) % n]);
    }
    return area / 2;
}
// ********************************************


// 凸性判定 (IsConvex) ************************
bool IsConvex(Polygon pol){
    int n = pol.size();
    for(int i = 0; i < n; i++){
        if(CCW(pol[i], pol[(i+1) % n], pol[(i+2) % n]) == -1) return false;
    }
    return true;
}
// ********************************************


// 多角形-点包含関係 (PolygonPointContainment)
int PolygonPointContainment(Polygon pol, Point p){
    enum {OUT, ON, IN};
    bool in = false;
    int n = pol.size();
    for(int i = 0; i < n; i++){
        Point a = pol[i] - p, b = pol[(i+1) % n] - p;
        if(a.Y > b.Y) swap(a, b);
        if(a.Y <= 0 && 0 < b.Y && Cross(a, b) < 0) in = !in;
        if(Cross(a, b) == 0 && Dot(a, b) <= 0) return ON;
    }
    return in ? IN : OUT;
}
// ********************************************


//*********************************************
//          凸多角形(ConvexPolygon)           *
//*********************************************

// 凸包[θ<=180] (ConvexHull) *****************
Polygon ConvexHull(Polygon pol){
    int n = pol.size(), k = 0;
    sort(pol.begin(), pol.end());
    Polygon ch(2*n);
    for(int i = 0; i < n; ch[k++] = pol[i++])
        while(k >= 2 && CCW(ch[k-2], ch[k-1], pol[i]) == -1) --k;

    for(int i = n-2, t = k+1; i >= 0; ch[k++] = pol[i--])
        while(k >= t && CCW(ch[k-2], ch[k-1], pol[i]) == -1) --k;

    ch.resize(k-1);
    return ch;
}
// ********************************************

// 凸包[θ<180] (ConvexHull2) *****************
Polygon ConvexHull2(Polygon pol){
    int n = pol.size(), k = 0;
    sort(pol.begin(), pol.end());
    Polygon ch(2*n);
    for(int i = 0; i < n; ch[k++] = pol[i++])
        while(k >= 2 && CCW(ch[k-2], ch[k-1], pol[i]) <= 0) --k;

    for(int i = n-2, t = k+1; i >= 0; ch[k++] = pol[i--])
        while(k >= t && CCW(ch[k-2], ch[k-1], pol[i]) <= 0) --k;

    ch.resize(k-1);
    return ch;
}
// ********************************************

// 凸多角形の直径 (ConvexPolygonDiameter) *****
pair<int, int> ConvexPolygonDiameter(Polygon cpol){
    int n = cpol.size();
    int is = 0, js = 0;
    for (int i = 1; i < n; ++i) {
        if (imag(cpol[i]) > imag(cpol[is])) is = i;
        if (imag(cpol[i]) < imag(cpol[js])) js = i;
    }
    LD maxd = norm(cpol[is]-cpol[js]);

    int i, maxi, j, maxj;
    i = maxi = is;
    j = maxj = js;
    do{
        if(Cross(cpol[(i+1)%n] - cpol[i], cpol[(j+1)%n] - cpol[j]) >= 0) j = (j+1) % n;
        else i = (i+1) % n;
        if(norm(cpol[i]-cpol[j]) > maxd) {
            maxd = norm(cpol[i]-cpol[j]);
            maxi = i; maxj = j;
        }
    } while(i != is || j != js);
    return make_pair(maxi, maxj);
}
// ********************************************

// 凸多角形-点包含関係 (ConvexPolygonPointContainment)
int ConvexPolygonPointContainment(Polygon cpol, Point p){
    enum {OUT, ON, IN};
    int n = cpol.size();
    Point g = (cpol[0] + cpol[n/3] + cpol[2*n/3]) / 3.0L;
    int a = 0, b = n;
    while(a+1 < b){
        int c = (a + b) / 2;
        if(Cross(cpol[a]-g, cpol[c]-g) > 0){
            if(Cross(cpol[a]-g, p-g) > 0 && Cross(cpol[c]-g, p-g) < 0) b = c;
            else a = c;
        }else{
            if(Cross(cpol[a]-g, p-g) < 0 && Cross(cpol[c]-g, p-g) > 0) a = c;
            else b = c;
        }
    }
    b %= n;
    if(Cross(cpol[a]-p, cpol[b]-p) < 0) return OUT;
    if(Cross(cpol[a]-p, cpol[b]-p) > 0) return IN;
    return ON;
}
// ********************************************


/************** using variables ***************/
Polygon p;
int n;
int q;
/**********************************************/

signed main(){
    cin >> n;
    REP(i, n){
        LD xp, yp;
        cin >> xp >> yp;
        p.pb(Point(xp, yp));
    }
    pair<int, int> ans = ConvexPolygonDiameter(p);
    printf("%.9Lf\n", DistPP(p[ans.first], p[ans.second]));

}