Problem:
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Problem F:
There is No Alternative

ICPC (Isles of Coral Park City) consist of several beautiful islands.

The citizens requested construction of bridges between islands to resolve inconveniences of using boats between islands, and they demand that all the islands should be reachable from any other islands via one or more bridges.

The city mayor selected a number of pairs of islands, and ordered a building company to estimate the costs to build bridges between the pairs. With this estimate, the mayor has to decide the set of bridges to build, minimizing the total construction cost.

However, it is difficult for him to select the most cost-efficient set of bridges among those connecting all the islands. For example, three sets of bridges connect all the islands for the Sample Input 1. The bridges in each set are expressed by bold edges in Figure F.1.

Figure F.1. Three sets of bridges connecting all the islands for Sample Input 1

As the first step, he decided to build only those bridges which are contained in all the sets of bridges to connect all the islands and minimize the cost. We refer to such bridges as no alternative bridges. In Figure F.2, no alternative bridges are drawn as thick edges for the Sample Input 1, 2 and 3.

Figure F.2. No alternative bridges for Sample Input 1, 2 and 3

Write a program that advises the mayor which bridges are no alternative bridges for the given input.

Input

The input consists of a single test case.

$N$ $M$

$S_1$ $D_1$ $C_1$

.

.

.

$S_M$ $D_M$ $C_M$

The first line contains two positive integers $N$ and $M$. $N$ represents the number of islands and each island is identified by an integer 1 through $N$. $M$ represents the number of the pairs of islands between which a bridge may be built.

Each line of the next $M$ lines contains three integers $S_i$, $D_i$ and $C_i$ ($1 \leq i \leq M$) which represent that it will cost $C_i$ to build the bridge between islands $S_i$ and $D_i$. You may assume $3 \leq N \leq 500$, $N − 1 \leq M \leq min(50000, N(N − 1)/2)$, $1 \leq S_i < D_i \leq N$, and $1 \leq C_i \leq 10000$. No two bridges connect the same pair of two islands, that is, if $i \ne j$ and $S_i = S_j$, then $D_i \ne D_j$ . If all the candidate bridges are built, all the islands are reachable from any other islands via one or more bridges.

Output

Output two integers, which mean the number of no alternative bridges and the sum of their construction cost, separated by a space.

Sample Input 1

4 4
1 2 3
1 3 3
2 3 3
2 4 3

Sample Output 1

1 3

Sample Input 2

4 4
1 2 3
1 3 5
2 3 3
2 4 3

Sample Output 2

3 9

Sample Input 3

4 4
1 2 3
1 3 1
2 3 3
2 4 3

Sample Output 3

2 4

Sample Input 4

3 3
1 2 1
2 3 1
1 3 1

Sample Output 4

0 0

Solution in C# (judge verdict: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using Debug = System.Diagnostics.Debug;
using SB = System.Text.StringBuilder;
using System.Numerics;
using static System.Math;
using P = System.Numerics.Complex;
using Number = System.Int64;
namespace Program {
    public class Solver {
        Random rnd = new Random(0);
        public void Solve() {
            var n = ri;
            var m = ri;
            var E = Enumerate(10050, x => new List<int>());
            var from = new int[m];
            var to = new int[m];
            for (int i = 0; i < m; i++)
            {
                from[i] = ri - 1;
                to[i] = ri - 1;
                E[ri].Add(i);
            }
            var set = new DisjointSet(n);
            var cnt = 0;
            Action<List<int>, int, int> dfs = null;
            dfs = (a, l, r) =>
            {
                Debug.WriteLine($"{l} {r}");
                if (r - l == 1)
                {
                    var i = a[l];
                    if (set.Find(from[i]) != set.Find(to[i]))
                    {
                        Debug.WriteLine(i);
                        cnt++;
                    }
                    return;
                }
                var lr = (l + r) / 2;
                var shot = set.a.Count;
                for (int u = lr; u < r; u++)
                {
                    var i = a[u];
                    set.Unite(from[i], to[i]);
                }
                dfs(a, l, lr);
                set.Back(shot);
                for (int u = l; u < lr; u++)
                {
                    var i = a[u];
                    set.Unite(from[i], to[i]);
                }
                dfs(a, lr, r);
                set.Back(shot);
            };
            var ans = 0;
            var ans2 = 0;
            for (int i = 0; i < 10050; i++)
                if (E[i].Count > 0)
                {
                    cnt = 0;
                    dfs(E[i], 0, E[i].Count);
                    foreach (var id in E[i]) set.Unite(from[id], to[id]);
                    ans += cnt;
                    ans2 += i * cnt;
                }
            Console.WriteLine($"{ans} {ans2}");
        }


        const long INF = 1L << 60;
        int ri { get { return sc.Integer(); } }
        long rl { get { return sc.Long(); } }
        double rd { get { return sc.Double(); } }
        string rs { get { return sc.Scan(); } }
        public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

        static T[] Enumerate<T>(int n, Func<int, T> f) {
            var a = new T[n];
            for (int i = 0; i < n; ++i) a[i] = f(i);
            return a;
        }
        static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
    }
}

#region main
static class Ex {
    static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
    static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
        return string.Join(st, ie);
    }
    static public void Main() {
        Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = false });
        var solver = new Program.Solver();
        solver.Solve();
        Console.Out.Flush();
    }
}
#endregion
#region Ex
namespace Program.IO {
    using System.IO;
    using System.Text;
    using System.Globalization;

    public class Printer: StreamWriter {
        public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
        public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
    }

    public class StreamScanner {
        public StreamScanner(Stream stream) { str = stream; }

        public readonly Stream str;
        private readonly byte[] buf = new byte[1024];
        private int len, ptr;
        public bool isEof = false;
        public bool IsEndOfStream { get { return isEof; } }

        private byte read() {
            if (isEof) return 0;
            if (ptr >= len)
            {
                ptr = 0;
                if ((len = str.Read(buf, 0, 1024)) <= 0)
                {
                    isEof = true;
                    return 0;
                }
            }
            return buf[ptr++];
        }

        public char Char() {
            byte b = 0;
            do b = read(); while ((b < 33 || 126 < b) && !isEof);
            return (char)b;
        }
        public string Scan() {
            var sb = new StringBuilder();
            for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
            return sb.ToString();
        }
        public string ScanLine() {
            var sb = new StringBuilder();
            for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
            return sb.ToString();
        }
        public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
        public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
        public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
    }
}

#endregion

class DisjointSet {
    int[] par;
    int[] rank;
    public List<KeyValuePair<int, int>> a = new List<KeyValuePair<int, int>>();
    public DisjointSet(int n) {
        rank = new int[n];
        par = new int[n];
        for (int i = 0; i < n; i++)
            par[i] = i;
    }
    public int Find(int i) {
        if (par[i] == i) return i;
        else return Find(par[i]);
    }
    public void Unite(int i, int j) {
        i = Find(i);
        j = Find(j);
        if (i == j) return;
        if (rank[i] > rank[j]) { var tmp = i; i = j; j = tmp; }
        a.Add(new KeyValuePair<int, int>(i, rank[i]));
        a.Add(new KeyValuePair<int, int>(j, rank[j]));
        if (rank[i] == rank[j]) rank[i]++;
        par[j] = i;
    }
    public void Back(int pos) {
        while (a.Count > pos)
        {
            var kv = a[a.Count - 1];
            var i = kv.Key;
            var r = kv.Value;
            par[i] = i; rank[i] = r;
            a.RemoveAt(a.Count - 1);
        }
    }
}