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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;

namespace ABC139F
{
    class Program
    {
        public static void Solve(Input input)
        {
            var n = input.NextInt();
            var x = new int[n];
            var y = new int[n];

            for (int i = 0; i < n; i++)
            {
                x[i] = input.NextInt();
                y[i] = input.NextInt();
            }

            var dx = new long[8];
            var dy = new long[8];
            for (int i = 0; i < n; i++)
            {
                // 第一象限方向で値が増加するもの
                if (x[i] >= 0 && y[i] >= 0 || (y[i] < 0 && x[i] > -y[i]) || (x[i] < 0 && y[i] > -x[i]))
                {
                    dx[0] += x[i];
                    dy[0] += y[i];
                }
                if (x[i] <= 0 && y[i] >= 0 || (y[i] < 0 && x[i] < y[i]) || (x[i] > 0 && y[i] > x[i]))
                {
                    dx[1] += x[i];
                    dy[1] += y[i];
                }
                if (x[i] <= 0 && y[i] <= 0 || (y[i] > 0 && -x[i] > y[i]) || (x[i] > 0 && -y[i] > x[i]))
                {
                    dx[2] += x[i];
                    dy[2] += y[i];
                }
                if (x[i] >= 0 && y[i] <= 0 || (y[i] > 0 && x[i] > y[i]) || (x[i] < 0 && y[i] < x[i]))
                {
                    dx[3] += x[i];
                    dy[3] += y[i];
                }

                // +X方向で値が増加するもの
                if (x[i] >= 0)
                {
                    dx[4] += x[i];
                    dy[4] += y[i];
                }
                if (y[i] >= 0)
                {
                    dx[5] += x[i];
                    dy[5] += y[i];
                }
                if (x[i] <= 0)
                {
                    dx[6] += x[i];
                    dy[6] += y[i];
                }
                if (y[i] <= 0)
                {
                    dx[7] += x[i];
                    dy[7] += y[i];
                }
            }

            var max = long.MinValue;
            for (int i = 0; i < dx.Length; i++)
            {
                max = Math.Max(max, dx[i] * dx[i] + dy[i] * dy[i]);
            }

            Console.WriteLine($"{Math.Sqrt(max):F30}");
        }

        #region Main
        public static void Main(string[] args)
        {
            // 出力が少ないときはこれをセットする方が時間かかるけれど
            // そのときにTLEするのはアルゴリズムが悪いせいだし、まあ良しとする
            var needsFlushOutput = true;
            if (needsFlushOutput)
            {
                // 細かく出力しないようにする
                var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
                Console.SetOut(sw);
            }

            // 通常は引数無しで、ファイルから読み出したい場合はpath指定する
            var input = new Input();

            // 仮想的に標準入力をセットする
            // テスト入力したまま提出することがままあるので……
#if DEBUG
            //input.SetText("3 5 5 5 5 -10 5");
            //input.SetText("2 -1 1 -1 1");
            //input.SetText("2 -1 -1 -1 -1");
            //input.SetText("2 1 -1 1 -1");
#endif

            Solve(input);

            Console.Out.Flush();
        }
        #endregion

        #region Competitive Template
#pragma warning disable CS0414
        static readonly int MOD = (int)1e9 + 7;
        static readonly int[] dx = { 1, 0, -1, 0 };
        static readonly int[] dy = { 0, 1, 0, -1 };
        /// <summary> 左上が原点 </summary>
        static readonly char[] dir = { 'R', 'U', 'L', 'D' };
#pragma warning restore CS0414

        public class Input
        {
            // 変な改行コードがたまに混じっているので、一応セパレート指定する
            // スペース単独指定の方がもちろん早い
            static readonly char[] separator = { ' ', '\r', '\n' };
            readonly StreamReader sr;
            readonly Queue<string> queue;

            /// <summary>
            /// 特定のファイルから読み出したい場合はpath指定する
            /// </summary>
            public Input(string path = "")
            {
                queue = new Queue<string>();

                if (string.IsNullOrEmpty(path)) { sr = new StreamReader(Console.OpenStandardInput()); }
                else { sr = new StreamReader(path); }
            }

            /// <summary>
            /// 入力予約
            /// </summary>
            public void SetText(IEnumerable<string> items)
            {
                foreach (var item in items)
                    SetText(item);
            }

            /// <summary>
            /// 入力予約
            /// </summary>
            public bool SetText(string s)
            {
                if (string.IsNullOrEmpty(s)) return false;

                foreach (var elem in s.Trim().Split(separator, StringSplitOptions.RemoveEmptyEntries))
                    queue.Enqueue(elem);

                return true;
            }

            /// <summary>
            /// 要素が存在するか
            /// </summary>
            public bool Any() => queue.Any() || Read();

            /// <summary>
            /// 内部queueに入力からの値をsplitして格納する
            /// </summary>
            bool Read()
            {
                if (!SetText(sr.ReadLine())) return false;

                if (!queue.Any()) return Read();

                return queue.Any();
            }

            /// <summary>
            /// 次のstringを一つ読み込む
            /// </summary>
            public string Next()
            {
                if (!queue.Any() && !Read()) return "";

                return queue.Dequeue();
            }

            /// <summary>
            /// 指定個数queueにたまるまでenqueueし続ける
            /// </summary>
            bool Accumulate(int n)
            {
                while (queue.Count() < n)
                    if (!Read()) return false;

                return true;
            }

            public int NextInt() => int.Parse(Next());
            public long NextLong() => long.Parse(Next());
            public double NextDouble() => double.Parse(Next());

            /// <summary>
            /// n個の要素をparseして、それぞれにoffsetをaddした配列を返す
            /// </summary>
            T[] NextT<T>(int n, T offset, Func<string, T> parse, Func<T, T, T> add)
            {
                if (!Accumulate(n)) return null;

                var a = new T[n];

                for (int i = 0; i < n; i++)
                    a[i] = add(parse(queue.Dequeue()), offset);

                return a;
            }

            public string[] Next(int n) => NextT(n, "", x => x, (x, y) => x);
            public int[] NextInt(int n, int offset = 0) => NextT(n, offset, int.Parse, (x, y) => x + y);
            public long[] NextLong(int n, long offset = 0) => NextT(n, offset, long.Parse, (x, y) => x + y);
            public double[] NextDouble(int n, double offset = 0.0) => NextT(n, offset, double.Parse, (x, y) => x + y);
        }

        static class Utils
        {
            public static T Max<T>(params T[] objs) => objs.Max();
            public static T Min<T>(params T[] objs) => objs.Min();

            /// <summary>
            /// vでfillされたT[d1][d2]配列を作成する
            /// </summary>
            public static T[][] Create2DArray<T>(int d1, int d2, T v)
            {
                return
                    Enumerable.Repeat(0, d1).Select(_ =>
                    Enumerable.Repeat(v, d2).ToArray()).ToArray();
            }

            /// <summary>
            /// vでfillされたT[d1][d2][d3]配列を作成する
            /// </summary>
            public static T[][][] Create3DArray<T>(int d1, int d2, int d3, T v)
            {
                return
                    Enumerable.Repeat(0, d1).Select(_ =>
                    Enumerable.Repeat(0, d2).Select(__ =>
                    Enumerable.Repeat(v, d3).ToArray()).ToArray()).ToArray();
            }
        }
        #endregion
    }
}