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>
typedef long long ll;

using namespace std;
using P = pair<int, int>;
using PP = pair<int, P>;

int N, M;
struct edge {
	P to;
	int cost;
};

template <typename T1, typename T2, typename T3>
int bfs(T1& G, T2& d, T3& used, P s) {
	deque<PP> deq;
	deq.push_back(PP(0, s));

	while (!deq.empty()) {
		PP q = deq.front();
		deq.pop_front();
		int cost = q.first;
		P v = q.second;
		if (used[v]) { continue; }
		used[v] = true;
		if (v == P(N - 1, -1)) { return cost; }

		for (int i = 0; i < G[v].size(); ++i) {
			if (used[G[v][i].to]) { continue; }
			if (G[v][i].cost == 0) {
				deq.push_front(PP(cost, G[v][i].to));
			} else {
				deq.push_back(PP(cost + 1, G[v][i].to));
			}
		}
	}
	return -1;
}

int main() {
	const int INF = 1e9;
	cin >> N >> M;
	map<P, vector<edge>> G;
	for (int i = 0; i < M; ++i) {
		int p, q, c;
		cin >> p >> q >> c;
		--p; --q;

		G[P(p, c)].push_back({P(q, c), 0});
		G[P(p, c)].push_back({P(p, -1), 0});
		G[P(p, -1)].push_back({P(q, c), 1});
		G[P(q, c)].push_back({P(p, c), 0});
		G[P(q, c)].push_back({P(q, -1), 0});
		G[P(q, -1)].push_back({P(p, c), 1});
	}

	map<P, int> d;
	map<P, bool> used;
	for (auto it = G.begin(); it != G.end(); ++it) {
		d[it->first] = INF;
		used[it->first] = false;
	}

	int res = bfs(G, d, used, P(0, -1));
	cout << res << endl;

	return 0;
}