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: Runtime Error)

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>

#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; ++i)
#define repp(i, m, n) for(ll i = m, i##_len = (n); i < i##_len; ++i)
#define all(x) (x).begin(), (x).end()
template <class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1; } return 0; }
template <class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }
template <class T> T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair <int,int> P;
typedef long double ld;

int main(void)
{
	int n, m;
	cin >> n >> m;
	vector<P> sta[n]; // (行先,路線)
	rep (i, m) {
		int p, q, c;
		cin >> p >> q >> c;
		sta[p - 1].push_back(make_pair(q - 1, c));
	}
	deque<int> deq; // 0-1BFSの管理用
	int cost[n] = {}, cmp = -1;
	vector<int> cmps[n]; // 最短経路で到達するのに使った路線
	repp (i, 1, n) cost[i] = INT32_MAX;
	deq.push_front(0);
	while (deq.size() > 0) {
		int now = deq.front();
		deq.pop_front();
		for (auto des : sta[now]) {
			if (any_of(cmps[n].begin(), cmps[n].end(), [&](int x) { return x == des.second; })) {
				if (cost[des.first] == INT32_MAX) {
					cost[des.first] = cost[now];
					deq.push_front(des.first);
				}
				if (cost[des.first] == cost[now]) {
					cmps[des.first].push_back(des.second);
				}
			} else {
				if (cost[des.first] == INT32_MAX) {
					cost[des.first] = cost[now] + 1;
					deq.push_back(des.first);
				}
				if (cost[des.first] == cost[now] + 1) {
					cmps[des.first].push_back(des.second);
				}
			}
		}
	}
	// rep (i, n) cout << cost[i] << " "; cout << endl;
	cout << (cost[n - 1] == INT32_MAX ? -1 : cost[n - 1]) << endl;
	return 0;
}