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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
int N, M, Q, P, C;

class Node {
public:
	int station;
	int cost;
	int company;
	Node* next;

	Node(int _s, int _cos, int _com) {
		station = _s;
		cost = _cos;
		company = _com;
		next = NULL;
	}
};

class Queue {
public:
	Node *head;
	Node *tail;

	Queue() {
		head = NULL;
		tail = NULL;
	}

	Node dequeue() {
		Node n = *head;
		if (head != NULL) {
			head = (*head).next;
		}
		if (head == NULL) {
			tail = NULL;
		}
		return n;
	}

	void enqueue(Node n) {
		if (head == NULL) {
			head = &n;
			tail = &n;
		} else {
			(*tail).next = &n;
			tail = &n;
		}
	}

	bool isEmpty() {
		return (head == NULL);
	}
};

int main() {
	cin >> N >> M;
	vector<vector<pair<int, int> > > rosen(N+1);
	vector<int> chk(N+1, 0);

	while (M--) {
		cin >> Q >> P >> C;
		pair<int, int> p1(P, C);
		pair<int, int> p2(Q, C);
		rosen[Q].push_back(p1);
		rosen[P].push_back(p2);
	}
	Queue q1, q2;
	Node n(1, 0, 0);
	q2.enqueue(n);
	int cost = 1;
	bool isFinish = false;
	int idx = 0;
	while (true) {
		vector<bool> chk2(N+1, false);
		chk2[idx] = true;
		while (!q1.isEmpty()) {
			Node n1 = q1.dequeue();
			chk2[n1.station] = true;
			for (auto r: rosen[n1.station]) {
				cost = n1.cost + ((n1.company == r.second) ? 0 : 1);
				if (r.first == N) {
					isFinish = true;
					break;
				}
				if ((chk[r.first] > 0 && chk[r.first] < cost) || chk2[r.first]) continue;
				if (n1.company == r.second) {
					chk2[r.first] = true;
				}
				chk[r.first] = cost;
				Node _n(r.first, cost, r.second);
				if (n1.company == r.second) {
					q1.enqueue(_n);
				} else {
					q2.enqueue(_n);
				}
			}
			if (isFinish) break;
		}
		if (isFinish || q2.isEmpty()) {
			break;
		}
		Node n2 = q2.dequeue();
		idx = n2.station;
 		for (auto r: rosen[n2.station]) {
 			cost = n2.cost + ((n2.company == r.second) ? 0 : 1);
			if (r.first == N) {
				isFinish = true;
				break;
			}
			if (chk[r.first] > 0 && chk[r.first] < cost) continue;
			chk[r.first] = cost;
			Node _n(r.first, cost, r.second);
			if (n2.company == r.second) {
				q1.enqueue(_n);
			} else {
				q2.enqueue(_n);
			}
		}
		if (isFinish) break;
	}

	if (!isFinish) cost = -1;
	cout << cost << endl;
}