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:
// /usr/local/bin/g++ Main.cpp -std=c++14
#include <bits/stdc++.h>

#define INF 4000001
using namespace std;

int n, m;
struct edge {int to, co;};
vector<edge> g[100010];
int dis[100010];
set<int> disc[100010];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m;
	for (int i = 0; i < m; i++) {
		int a, b, c;
		cin >> a >> b >> c;
		g[a].push_back({b, c});
		g[b].push_back({a, c});
	}

	dis[1] = 0;
	for (int i = 2; i <= n; i++) {
		dis[i] = INF;
	}
	priority_queue<pair<int, int> > q;
	q.push({-dis[1], 1});
	while(q.size()) {
		int now = q.top().second;
		int d = -q.top().first;
		q.pop();
		if (dis[now] < d) continue;
		for (edge j : g[now]) {
			int nd = dis[now];
			if (!disc[now].count(j.co)) nd++;
			if (nd < dis[j.to]) {
				dis[j.to] = nd;
				disc[j.to].clear();
				disc[j.to].insert(j.co);
				q.push({-dis[j.to], j.to});
			} else if (nd == dis[j.to]) {
				disc[j.to].insert(j.co);
			}
		}
	}

	if (dis[n] == INF) {
		cout << -1 << endl;
	} else {
		cout << dis[n] << endl;
	}
}