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>

const int MAXN = 100010;
int dis[MAXN];
std::map<int, std::vector<int> > es[MAXN];
std::set<int> vis[MAXN];
std::vector<int> upd, tmp;
void extend(int u, int c, int s) {
	if (s > dis[u] || vis[u].count(c)) return ;
	vis[u].insert(c);
	if (s < dis[u]) dis[u] = s, upd.push_back(u);
	for (auto t : es[u][c]) extend(t, c, s);
}
int n, m;
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	std::cin >> n >> m;
	for (int i = 1; i <= m; ++i) {
		int u, v, w;
		std::cin >> u >> v >> w;
		es[u][w].push_back(v);
		es[v][w].push_back(u);
	}
	memset(dis, 0x3f, sizeof dis);
	upd.push_back(1); dis[1] = 0;
	int S = 0;
	while (!upd.empty()) {
		++S;
		std::swap(upd, tmp);
		for (auto u : tmp)
			for (auto t : es[u])
				for (auto v : t.second)
					extend(v, t.first, S);
		tmp.clear();
	}
	std::cout << (dis[n] <= n ? dis[n] : -1) << std::endl;
	return 0;
}