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 <map>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int N, M, p[200009], q[200009], c[200009], dist[300009]; vector<int> g[300009]; map<pair<int, int>, vector<int> > d;
void add_edge(int a, int b) {
	g[a].push_back(b);
	g[b].push_back(a);
}
int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		cin >> p[i] >> q[i] >> c[i]; p[i]--, q[i]--;
		d[make_pair(p[i], c[i])].push_back(i);
		d[make_pair(q[i], c[i])].push_back(i);
		add_edge(p[i], i + N);
		add_edge(q[i], i + N);
	}
	for (pair<pair<int, int>, vector<int> > i : d) {
		for (int j = 1; j < i.second.size(); j++) {
			add_edge(i.second[j - 1] + N, i.second[j] + N);
		}
	}
	fill(dist, dist + N + M, 1012345678); dist[0] = 0;
	priority_queue<pair<int, int> > que; que.push(make_pair(0, 0));
	while (!que.empty()) {
		int u = que.top().second; que.pop();
		for (int i : g[u]) {
			int ds = dist[u] + (u >= N && i >= N ? 0 : 1);
			if (dist[i] > ds) {
				dist[i] = ds;
				que.push(make_pair(-ds, i));
			}
		}
	}
	cout << (dist[N - 1] == 1012345678 ? -1 : dist[N - 1] / 2) << endl;
	return 0;
}