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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#pragma region
#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <list>
#include <iomanip>
#include <cstdint>
#include <bitset>

using namespace std;
typedef long long ll;
//#define rep(i, s, e) for (int(i) = (s); (i) < (e); ++(i))
#define rep(i, e) for (int(i) = 0; (i) < (e); ++(i))
#define rrep(i, s) for (int(i) = (s) - 1; (i) >= 0; --(i))
#define all(x) x.begin(),x.end()

#pragma region UnionFind
struct UnionFind
{
	vector<int> par;

	UnionFind(int n) : par(n, -1) {}
	void init(int n) { par.assign(n, -1); }

	int root(int x)
	{
		if (par[x] < 0) return x;
		else return par[x] = root(par[x]);
	}

	bool issame(int x, int y)
	{
		return root(x) == root(y);
	}

	bool merge(int x, int y)
	{
		x = root(x); y = root(y);
		if (x == y) return false;
		if (par[x] > par[y]) swap(x, y);
		par[x] += par[y];
		par[y] = x;
		return true;
	}

	int size(int x)
	{
		return -par[root(x)];
	}
};
#pragma endregion
#pragma region GCD
int gcd(int a, int b)
{
	if (a%b == 0)return b;
	return gcd(b, a%b);
}
#pragma endregion
#pragma region chmin
template<class T> inline bool chmin(T& a, T b)
{
	if (a > b)
	{
		a = b;
		return true;
	}
	return false;
}
#pragma endregion
#pragma region chmax
template<class T> inline bool chmax(T& a, T b)
{
	if (a < b)
	{
		a = b;
		return true;
	}
	return false;
}
#pragma endregion
#pragma endregion


int main()
{
	int n, m; cin >> n >> m;
	vector<vector<pair<int, int>>> v(n);
	rep(i, m)
	{
		int p, q, c; cin >> p >> q >> c;
		--p, --q;
		v[p].push_back(make_pair(q, c));
		v[q].push_back(make_pair(p, c));
	}
	vector<int> cost(n, 1e6);
	cost[0] = 0;
	deque<pair<int, int>> d;
	d.push_back(make_pair(0, 1e6));
	while (!d.empty())
	{
		int station, line;
		tie(station, line) = d.front();
		d.pop_front();
		for (auto p : v[station])
		{
			int next_station = p.first;
			int next_line = p.second;
			bool change_lines = !(next_line == line);
			if (change_lines && chmin(cost[next_station], cost[station] + 1))d.push_back(make_pair(next_station, next_line));
			if (!change_lines && chmin(cost[next_station], cost[station]))d.push_front(make_pair(next_station, next_line));
		}
	}
	if (cost[n - 1] == 1e6)cout << -1 << endl;
	else cout << cost[n - 1] << endl;
}