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;

const int INF = 1e9;

const int MAXN = 1.1e5;
const int MAXM = 2.1e5;
int N, M;
map<int, vector<int>> adj[MAXN];

map<int, int> dist[MAXN];

bool update(int v, int c, int d) {
	if (!dist[v].count(c) || dist[v][c] > d) {
		dist[v][c] = d;
		return true;
	} else {
		return false;
	}
}

int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	cin >> N >> M;
	for (int e = 0; e < M; e++) {
		int p, q, c; cin >> p >> q >> c;
		adj[p][c].push_back(q);
		adj[q][c].push_back(p);
	}

	deque<tuple<int, int, int>> que;
	que.emplace_back(0, 1, 0);
	dist[1][0] = 0;

	while (!que.empty()) {
		int d, v, c; tie(d, v, c) = que.front();
		que.pop_front();
		if (d != dist[v][c]) continue;

		if (v == N) {
			cout << d << '\n';
			exit(0);
		}

		if (c == 0) {
			for (auto& p : adj[v]) {
				if (update(v, p.first, d+1)) {
					que.emplace_back(d+1, v, p.first);
				}
			}
		} else {
			if (update(v, 0, d)) {
				que.emplace_front(d, v, 0);
			}
			for (auto& nv : adj[v][c]) {
				if (update(nv, c, d)) {
					que.emplace_front(d, nv, c);
				}
			}
		}
	}

	cout << -1 << '\n';

	return 0;
}