Problem:
Score : 600 points

Problem Statement

Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.

The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.

You can change trains at a station where multiple lines are available.

The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.

Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq 2×10^5

1 \leq p_i \leq N (1 \leq i \leq M)

1 \leq q_i \leq N (1 \leq i \leq M)

1 \leq c_i \leq 10^6 (1 \leq i \leq M)

p_i \neq q_i (1 \leq i \leq M)

Input

The input is given from Standard Input in the following format:

N M
p_1 q_1 c_1
:
p_M q_M c_M

Output

Print the minimum required fare. If it is impossible to get to station N by subway, print -1 instead.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;

class MainClass
{
	public static void Main()
	{
		new P().Solve();
	}
}
class P
{
	const int Inf = 1073741789;
	int N, M, V;
	int[] us, vs, ws;
	List<int>[] es;
	HashSet<int> dead;
	int[] F()
	{
		return Console.ReadLine().Split().Select(_ => int.Parse(_)).ToArray();
	}
	public void Solve()
	{
		var I = F();
		N = I[0];
		M = I[1];
		us = new int[M];
		vs = new int[M];
		ws = new int[M];
		var cs = new SortedSet<int>[N];
		for (var i = 0; i < N; i++) cs[i] = new SortedSet<int>(new int[]{ -1 });
		for (var i = 0; i < M; i++)
		{
			I = F();
			var a = us[i] = I[0] - 1;
			var b = vs[i] = I[1] - 1;
			var c = ws[i] = I[2] - 1;
			cs[a].Add(c);
			cs[b].Add(c);
		}
		var sep = new int[N + 1];
		for (var i = 0; i < N; i++) sep[i + 1] = sep[i] + cs[i].Count;
		dead = new HashSet<int>(sep);
		V = sep[N];
		var index = new Dictionary<int, int>[N];
		for (int i = 0, j = 0; i < N; i++)
		{
			index[i] = new Dictionary<int, int>();
			foreach (var c in cs[i]) index[i].Add(c, j++);
		}
		es = new List<int>[V];
		for (var i = 0; i < V; i++) es[i] = new List<int>();
		for (var i = 0; i < M; i++)
		{
			int a = us[i], b = vs[i], c = ws[i], u = index[a][c], v = index[b][c];
			es[u].Add(v);
			es[v].Add(u);
		}
		for (var i = 0; i < N; i++) for (var j = sep[i] + 1; j < sep[i + 1]; j++)
			{
				es[j].Add(sep[i]);
				es[sep[i]].Add(j);
			}
		var d = DijkstraFrom(0);
		Console.WriteLine(d[sep[N - 1]] == Inf ? -1 : d[sep[N - 1]]);
	}
	int[] DijkstraFrom(int from)
	{
		var d = new int[V];
		for (var i = 0; i < V; i++) d[i] = Inf;
		var queue = new PriorityQueue();
		d[from] = 0;
		queue.Enqueue(new Pair(from, 0));
		while (!queue.IsEmpty)
		{
			var p = queue.Dequeue();
			int u = p.First, cost = dead.Contains(u) ? 1 : 0;
			if (d[u] < p.Second) continue;
			foreach (var v in es[u])
			{
				int tmp = d[u] + cost;
				if (d[v] > tmp) queue.Enqueue(new Pair(v, d[v] = tmp));
			}
		}
		return d;
	}
}
class PriorityQueue
{
	List<Pair> list = new List<Pair>();
	public int Count { get; private set; }
	public bool IsEmpty { get { return Count == 0; } }
	public void Enqueue(Pair x)
	{
		var pos = Count++;
		list.Add(x);
		while (pos > 0)
		{
			var p = (pos - 1) / 2;
			if (list[p].Second <= x.Second) break;
			list[pos] = list[p];
			pos = p;
		}
		list[pos] = x;
	}
	public Pair Dequeue()
	{
		var value = list[0];
		var x = list[--Count];
		list.RemoveAt(Count);
		if (Count == 0) return value;
		var pos = 0;
		while (pos * 2 + 1 < Count)
		{
			var a = 2 * pos + 1;
			var b = 2 * pos + 2;
			if (b < Count && list[b].Second < list[a].Second) a = b;
			if (list[a].Second >= x.Second) break;
			list[pos] = list[a];
			pos = a;
		}
		list[pos] = x;
		return value;
	}
}
class Pair
{
	public int First;
	public int Second;
	public Pair(int s = 0, int t = 0)
	{
		First = s;
		Second = t;
	}
}