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: Accepted)

#include "iostream"
#include "algorithm"
#include "string"
#include "vector"
#include "cmath"
#include "bitset"
#include "queue"
#include "functional"
#include "map"
#include "unordered_map"
#include "set"
#include "unordered_set"
#include "stack"
#include "tuple"

#define rep(n) for (int i = 0; i < n; ++i)
#define REP(n,i) for (int i = 0; i < n; ++i)
#define mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define nyan "(=^･ω･^=)"
#define mkp make_pair
#define mkt make_tuple
#define P pair<int,int>

typedef long long ll;
using namespace std;

int n, m, p, q, c;

namespace std {
	template<>
	class hash<P> {
	public:
		size_t operator()(const P&x)const {
			return hash<ll>()(x.first)*intmax + hash<ll>()(x.second);
		}
	};
}

map<P, vector<pair<P, int>>>v;
map<P, int>mp;
priority_queue<pair<int, P>, vector<pair<int, P>>, greater<pair<int, P>>>Q;

int main() {
	cin >> n >> m;
	rep(m) {
		cin >> p >> q >> c;
		v[mkp(p, c)].push_back(mkp(mkp(q, c), 0));
		v[mkp(q, c)].push_back(mkp(mkp(p, c), 0));
	}
	for (auto it = v.begin(); it != v.end(); ++it) {
		if (it->first.second != 0) {
			it->second.push_back(mkp(mkp(it->first.first, 0), 1));
			v[mkp(it->first.first, 0)].push_back(mkp(it->first, 0));
		}
	}
	mp[mkp(1, 0)] = 0;
	rep(v[mkp(1, 0)].size()) {
		Q.push(mkp(v[mkp(1, 0)][i].second, v[mkp(1, 0)][i].first));
	}
	while (!Q.empty()) {
		P Qsec = Q.top().second;
		int Qfst = Q.top().first;
		Q.pop();
		if (mp.find(Qsec)==mp.end()) {
			mp[Qsec] = Qfst;
			rep(v[Qsec].size()) {
				if (mp.find(v[Qsec][i].first)==mp.end()) {
					Q.push(mkp(Qfst + v[Qsec][i].second, v[Qsec][i].first));
				}
			}
		}
	}
	if (mp.find(mkp(n, 0)) == mp.end())cout << -1 << endl;
	else cout << mp[mkp(n, 0)] << endl;
	return 0;
}