Problem:
Score: 600 points

Problem Statement

E869120 is initially standing at the origin (0, 0) in a two-dimensional plane.

He has N engines, which can be used as follows:

When E869120 uses the i-th engine, his X- and Y-coordinate change by x_i and y_i, respectively. In other words, if E869120 uses the i-th engine from coordinates (X, Y), he will move to the coordinates (X + x_i, Y + y_i).

E869120 can use these engines in any order, but each engine can be used at most once. He may also choose not to use some of the engines.

He wants to go as far as possible from the origin.
Let (X, Y) be his final coordinates. Find the maximum possible value of \sqrt{X^2 + Y^2}, the distance from the origin.

Constraints

1 \leq N \leq 100

-1 \ 000 \ 000 \leq x_i \leq 1 \ 000 \ 000

-1 \ 000 \ 000 \leq y_i \leq 1 \ 000 \ 000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
x_2 y_2
: :
x_N y_N

Output

Print the maximum possible final distance from the origin, as a real value.
Your output is considered correct when the relative or absolute error from the true answer is at most 10^{-10}.

Sample Input 1

3
0 10
5 -5
-5 -5

Sample Output 1

10.000000000000000000000000000000000000000000000000

The final distance from the origin can be 10 if we use the engines in one of the following three ways:

Use Engine 1 to move to (0, 10).

Use Engine 2 to move to (5, -5), and then use Engine 3 to move to (0, -10).

Use Engine 3 to move to (-5, -5), and then use Engine 2 to move to (0, -10).

The distance cannot be greater than 10, so the maximum possible distance is 10.

Sample Input 2

5
1 1
1 0
0 1
-1 0
0 -1

Sample Output 2

2.828427124746190097603377448419396157139343750753

The maximum possible final distance is 2 \sqrt{2} = 2.82842....
One of the ways to achieve it is:

Use Engine 1 to move to (1, 1), and then use Engine 2 to move to (2, 1), and finally use Engine 3 to move to (2, 2).

Sample Input 3

5
1 1
2 2
3 3
4 4
5 5

Sample Output 3

21.213203435596425732025330863145471178545078130654

If we use all the engines in the order 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 5, we will end up at (15, 15), with the distance 15 \sqrt{2} = 21.2132... from the origin.

Sample Input 4

3
0 0
0 1
1 0

Sample Output 4

1.414213562373095048801688724209698078569671875376

There can be useless engines with (x_i, y_i) = (0, 0).

Sample Input 5

1
90447 91000

Sample Output 5

128303.000000000000000000000000000000000000000000000000

Note that there can be only one engine.

Sample Input 6

2
96000 -72000
-72000 54000

Sample Output 6

120000.000000000000000000000000000000000000000000000000

There can be only two engines, too.

Sample Input 7

10
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
17 18
19 20

Sample Output 7

148.660687473185055226120082139313966514489855137208

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>

#include <vector>
#include <map>
#include <algorithm>
#include <fstream>
#include<cstdio>
#include<iomanip>
#include<stack>
#include<queue>
#include<string>
#include <cstdlib>
#include <typeinfo>
#include <math.h>
#include <list>

#define REP(i, n) for(int i=0;i<n;i++)
#define REP2(i, s, n) for(int i=s;i<n;i++)
#define REP_1(i, n) for(int i=1;i<n+1;i++)
#define bitSearch(bit, n) for(int bit = 0; bit < (1 << n); bit++)
#define sz(x) int(x.size())
using namespace std;

template<class T>
void print(const T &value) {
    std::cout << value << std::endl;
}

void yesno(bool a) { if (a)cout << "yes" << endl; else cout << "no" << endl; }

void YesNo(bool a) { if (a)cout << "Yes" << endl; else cout << "No" << endl; }

void YESNO(bool a) { if (a)cout << "YES" << endl; else cout << "NO" << endl; }

typedef long long ll;
typedef unsigned long ul;
typedef long double ld;

template<class T>
inline bool chmin(T &a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

//template<class T>
//inline bool chmax(T &a, T b) {
//    if (a < b) {
//        a = b;
//        return true;
//    }
//    return false;
//}


ll INF = 1000001000;
ll mod = 1000000007;//10^9+7
ll power(ll x, ll n, ll mod) {//x^nをmodで割った余り
    ll res = 1;
    if (n > 0) {
        res = power(x, n / 2, mod);
        if (n % 2 == 0) res = (res * res) % mod;
        else res = (((res * res) % mod) * x) % mod;
    }
    return res;
}

int dx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int dy[8] = {0, -1, 1, 0, -1, 1, -1, 1};

struct Edge {
    int to, cost;

    Edge(int to, int cost) : to(to), cost(cost) {};
};

using Graph = vector<vector<int>>;//隣接リスト:G[i]にはiと隣接する頂点が入るよ。
using GraphW = vector<vector<Edge>>;//重み付きの隣接リスト
using P = pair<int, int>;//BFSで利用。queueに入れる。
using lP = pair<ll, ll>;
using PP = pair<int, P>; //dijkstraで利用,priority_queueに入れる。
using p_queue = priority_queue<int, vector<int>, greater<int>>;//小さい順

class UnionFindTree {

public:
    UnionFindTree(int size) : memberSize(size) {
        par.resize(size * sizeof(ll));
        rnk.resize(size * sizeof(ll));
        diff_weight.resize(size * sizeof(ll));
        REP(i, size) {
            par[i] = i;
            rnk[i] = 0;
            diff_weight[i] = 0;
        }
    }

    int memberSize;

    vector<int> par;
    vector<int> rnk;
    vector<ll> diff_weight;

//n個のnodeで初期化する関数
    void init(ll n) {
        REP(i, n) {
            par[i] = i;
            rnk[i] = 1;
        }
    }

//木の根を求める。この際，親は根に更新される。
    int find(ll x) {
        if (par[x] == x) { return x; }
        else {
            int r = find(par[x]);
            diff_weight[x] += diff_weight[par[x]];
            return par[x] = r;
        }
    }

//xとyが同じ集合に属するか判定
    bool same(ll x, ll y) {
        return find(x) == find(y);
    }

//xとyの属する集合を融合
    void unite(ll x, ll y) {
        x = find(x);
        y = find(y);
        if (x == y) { return; }

        if (rnk[x] < rnk[y]) {
            par[x] = y;
        } else {
            par[y] = x;
            if (rnk[x] == rnk[y]) { rnk[x]++; }
        }
    }

    //xの重み
    ll weight(ll x) {
        find(x);
        return diff_weight[x];
    }

    ll diff(ll x, ll y) {
        return weight(y) - weight(x);
    }

    //重みごと更新。weight(y)-weight(x)=wとなるようにする
    void merge(int x, int y, int w) {
        w += weight(x);
        w -= weight(y);
        x = find(x);
        y = find(y);
        if (x == y) { return; }
        if (rnk[x] < rnk[y]) {
            par[x] = y;
            diff_weight[x] = -w;
        } else {
            par[y] = x;
            diff_weight[y] = w;
            if (rnk[x] == rnk[y]) { rnk[x]++; }
        }
    }
};

priority_queue<P, vector<P>, greater<P>> que;

struct Coordinate {
    int x, y;

    Coordinate(int y, int x) : y(y), x(x) {};
};

ll gcd(ll a, ll b) {
    if (a < b) {
        ll tmp = a;
        a = b;
        b = tmp;
    }
    ll r = a % b;
    while (r != 0) {
        a = b;
        b = r;
        r = a % b;
    }
    return b;
}

bool chmax(ld &a, ld b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

//番号ズレ注意！！
int main() {
    ld x_xMyM = 0, y_xMyM = 0;
    ld x_xMym = 0, y_xMym = 0;
    ld x_xmyM = 0, y_xmyM = 0;
    ld x_xmym = 0, y_xmym = 0;
    ld x_yMxM = 0, y_yMxM = 0;
    ld x_yMxm = 0, y_yMxm = 0;
    ld x_ymxM = 0, y_ymxM = 0;
    ld x_ymxm = 0, y_ymxm = 0;
    ld N;
    cin >> N;
    REP(i, N) {
        ld x, y;
        cin >> x >> y;
        if (x > 0) {
            x_xMyM += x;
            x_xMym += x;
            y_xMyM += y;
            y_xMym += y;
        } else if (x < 0) {
            x_xmyM += x;
            x_xmym += x;
            y_xmyM += y;
            y_xmym += y;
        } else {
            if(y>0){x_xMyM+=x;x_xmyM+=x;y_xMyM+=y;y_xmyM+=y;}
            if(y<0){x_xMym+=x;x_xmym+=x;y_xMym+=y;y_xmym+=y;}
        }

        if (y > 0) {
            x_yMxM += x;
            x_yMxm += x;
            y_yMxM += y;
            y_yMxm += y;
        } else if (y < 0) {
            x_ymxM += x;
            x_ymxm += x;
            y_ymxM += y;
            y_ymxm += y;
        } else {
            if(x>0){x_yMxM+=x;x_ymxM+=x;y_yMxM+=y;y_ymxM+=y;}
            if(x<0){x_yMxm+=x;x_ymxm+=x;y_yMxm+=y;y_ymxm+=y;}
        }
    }
    ld ans=sqrt(x_xMyM*x_xMyM+y_xMyM*y_xMyM);
    chmax(ans,sqrt(x_xMym*x_xMym+y_xMym*y_xMym));
    chmax(ans,sqrt(x_xmym*x_xmym+y_xmym*y_xmym));
    chmax(ans,sqrt(x_xmyM*x_xmyM+y_xmyM*y_xmyM));
    chmax(ans,sqrt(x_yMxM*x_yMxM+y_yMxM*y_yMxM));
    chmax(ans,sqrt(x_yMxm*x_yMxm+y_yMxm*y_yMxm));
    chmax(ans,sqrt(x_ymxm*x_ymxm+y_ymxm*y_ymxm));
    chmax(ans,sqrt(x_ymxM*x_ymxM+y_ymxM*y_ymxM));
    cout << fixed << setprecision(40) << ans << endl;
}