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 <string>
#include <cstdlib>
#include <map>
#include <vector>
#include <algorithm>
#include <queue>
#include <functional>
using namespace std;
typedef long long ll;

int N, M;
#define INF 1e9
int status[100001], cost[100001];
vector<vector<pair<int, int>>>E;

void dijkstra(int s) {
	//初期化
	for (int i = 0; i < N; i++) {
		status[i] = 0;
		cost[i] = INF;
	}
	cost[s] = 0;
	status[s] = 1;//コスト最小化中

	priority_queue<pair<int, pair<int,int>>, vector<pair<int, pair<int,int>>>, greater<pair<int, pair<int,int>>>>PQ;
	PQ.push(make_pair(cost[s], make_pair(s,-1)));

	while (!PQ.empty()) {
		pair<int,int>u = PQ.top().second; PQ.pop();
		if (status[u.first] == -1)continue;

		status[u.first] = -1;//コスト最小化終了

		//uに隣接した点のコストを更新
		for (int i = 0; i < E[u.first].size(); i++) {
			int nu = E[u.first][i].first;
			if (status[nu] != -1) {
				if (cost[nu] > cost[u.first] + (E[u.first][i].second != u.second)) {
					cost[nu] = cost[u.first] + (E[u.first][i].second != u.second);
					status[i] = 1;//コスト最小化中
					PQ.push(make_pair(cost[nu], make_pair(nu,E[u.first][i].second)));
				}
			}
		}
	}
}

int main()
{
	cin >> N >> M;
	E.resize(N);
	for (int i = 0; i < M; i++) {
		int p, q, c; cin >> p >> q >> c;
		p--; q--;
		E[p].push_back(make_pair(q, c));
		E[q].push_back(make_pair(p, c));
	}
	dijkstra(0);
	int ans = cost[N - 1];
	if (ans == INF)ans = -1;
	cout << ans << endl;
    return 0;
}