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 <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <functional>
using namespace std;

typedef long long ll;


vector<pair<int, int>> edg[100003];
set<pair<int, int>> used;

int main() {
	int n, m; cin >> n >> m;
	for (int i = 0; i < m; ++i) {
		int p, q, c; cin >> p >> q >> c;
		--p; --q;
		edg[p].push_back({ q, c });
		edg[q].push_back({ p, c });
	}

	priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> que;
	que.push({ 0, {0, -1} });
	while (!que.empty()) {
		int d = que.top().first; pair<int, int> pos = que.top().second;
		que.pop();
		if (pos.first == n - 1) {
			cout << d << endl;
			return 0;
		}
		if (used.count(pos) == 1) continue;
		used.insert(pos);
		int t = pos.first, c = pos.second;
		if (c == -1) {
			for (int i = 0; i < (int)edg[t].size(); ++i) {
				if (used.count(edg[t][i]) == 1) continue;
				que.push({ d + 1, edg[t][i] });
			}
		}
		else {
			for (int i = 0; i < (int)edg[t].size(); ++i) {
				if (c == edg[t][i].second) {
					if (used.count({ t, -1 }) == 1) continue;
					que.push({ d,{t,-1} });
				}
				else {
					if (used.count(edg[t][i]) == 1) continue;
					que.push({ d, edg[t][i] });
				}
			}
		}
	}
	cout << -1 << endl;
	return 0;
}