Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
typedef pair<P,int> PI;
typedef pair<int, P> S;
map<P, vector<PI>> G;
map<P, int> dist;
int main() {
	int N, M;
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int p, q, c;
		cin >> p >> q >> c;
		p--; q--;
		G[P(p, c)].emplace_back(P(q, c), 0);
		G[P(q, c)].emplace_back(P(p, c), 0);
		G[P(p, -1)].emplace_back(P(p, c), 1);
		G[P(p, c)].emplace_back(P(p, -1), 0);
		G[P(q, -1)].emplace_back(P(q, c), 1);
		G[P(q, c)].emplace_back(P(q, -1), 0);
	}
	priority_queue<S, vector<S>, greater<S>> Q;
	Q.push(S(0, P(0, -1)));
	dist[P(0, -1)] = 0;
	while (!Q.empty()) {
		S s = Q.top(); Q.pop();
		if (dist[s.second] < s.first) continue;
		for (auto e : G[s.second]) {
			int cost = s.first + e.second;
			if (dist.find(e.first) == dist.end() || dist[e.first] > cost) {
				dist[e.first] = cost;
				Q.push(S(cost, e.first));
			}
		}
	}
	if (dist.find(P(N - 1, -1)) == dist.end()) {
		cout << -1 << endl;
	}
	else {
		cout << dist[P(N - 1, -1)] << endl;
	}
}