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 <bits/stdc++.h>
using namespace std;

int n, m;
vector <int> g[400005];
vector <pair <int, pair <int, int> > > e;

int find(vector <int> &dsu, int x) {
	if (dsu[x] != x)
		return dsu[x] = find(dsu, dsu[x]);
	return x;
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1, u, v, c; i <= m; i ++) {
		scanf("%d%d%d", &u, &v, &c);
		e.push_back(make_pair(c, make_pair(u, v)));
	}
	int cc = n + 1;
	sort(e.begin(), e.end());
	for (int c = 1, i = 0, j = 0; c <= 100000 && j < (int) e.size(); c ++, i = j) {
		vector <int> v;
		for (j = i; j < (int) e.size() && e[j].first == e[i].first; j ++) {
			v.push_back(e[j].second.first);
			v.push_back(e[j].second.second);
		}
		sort(v.begin(), v.end());
		v.resize(unique(v.begin(), v.end()) - v.begin());
		vector <int> dsu;
		for (int k = 0; k < (int) v.size(); k ++)
			dsu.push_back(k);
		for (int k = i; k < j; k ++) {
			int x = lower_bound(v.begin(), v.end(), e[k].second.first) - v.begin();
			int y = lower_bound(v.begin(), v.end(), e[k].second.second) - v.begin();
			dsu[find(dsu, x)] = find(dsu, y);
		}
		vector <int> v1;
		for (int k = 0; k < (int) v.size(); k ++) {
			if (dsu[k] == k) v1.push_back(k);
		}
		sort(v1.begin(), v1.end());
		for (int k = 0; k < (int) v.size(); k ++) {
			int u = v[k];
			int v = cc + lower_bound(v1.begin(), v1.end(), find(dsu, k)) - v1.begin();
			// printf("%d %d\n", u, v);
			g[u].push_back(v);
			g[v].push_back(u);
		}
		cc += (int) v1.size();
	}
	queue <int> que;
	vector <int> dis;
	for (int i = 0; i < cc; i ++)
		dis.push_back(-1);
	dis[1] = 0;
	que.push(1);
	while (!que.empty()) {
		int u = que.front(); que.pop();
		for (int i = 0; i < (int) g[u].size(); i ++) {
			int v = g[u][i];
			if (!~dis[v]) {
				dis[v] = dis[u] + 1;
				que.push(v);
			}
		}
	}
	printf("%d\n", !~dis[n] ? -1 : dis[n] / 2);
	return 0;
}