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: Time Limit Exceeded)

#include <bits/stdc++.h>
using std::cout;
using std::endl;
using std::cin;

std::vector<std::map<int, std::vector<std::pair<std::pair<int, int>, int>>>> g;
std::vector<std::set<int>> st;
int n, m;

int main(){
	cin >> n >> m; g.resize(n); st.resize(n);
	for(int i = 0; i < m; i++){
		int a, b, c; cin >> a >> b >> c;
		
		g[a - 1][c].push_back({{b - 1, c}, 0});
		g[b - 1][c].push_back({{a - 1, c}, 0});
		st[a - 1].insert(c);
		st[b - 1].insert(c);
	}
	
	for(int i = 0; i < n; i++){
		for(auto s : st[i]){
			g[i][s].push_back({{i, -1}, 0});
			g[i][-1].push_back({{i, s}, 1});	
		}
	}
	
	
	std::priority_queue<std::pair<int, std::pair<int, int>>> qu; qu.push({0, {0, -1}});
	std::vector<std::map<int, int>> min_cost(n, std::map<int, int>()); min_cost[0][-1] = 0;
	while(!qu.empty()){
		auto p = qu.top(); qu.pop();
		
		for(auto e : g[p.second.first][p.second.second]){
			if(min_cost[e.first.first][e.first.second] and p.first + e.second >= min_cost[e.first.first][e.first.second]) continue;
			min_cost[e.first.first][e.first.second] = p.first + e.second;
			qu.push({p.first + e.second, e.first});
		}
	}
	
	cout << (min_cost[n - 1][-1] ? min_cost[n - 1][-1] : -1) << endl;
	return 0;
}