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>

int main()
{
	constexpr int inf{1 << 30};
	int N, M;
	scanf("%d%d", &N, &M);
	std::vector<std::vector<std::pair<int, int>>> edges(N);
	for (int i{}; i < M; i++)
	{
		int p, q, c;
		scanf("%d%d%d", &p, &q, &c);
		p--; q--;
		edges[p].push_back({q, c});
		edges[q].push_back({p, c});
	}

	std::vector<int> distance(N, inf);
	distance[0] = 0;
	std::queue<int> bfs;
	bfs.push(0);
	std::vector<std::set<int>> usedCompany(N);
	std::vector<bool> inQueue(N);
	inQueue[0] = true;

	while (!bfs.empty())
	{
		int now{bfs.front()};
		bfs.pop();
		inQueue[now] = false;
		for (auto& next: edges[now])
		{
			int dist{distance[now]};
			if (usedCompany[now].find(next.second) == usedCompany[now].end()) dist++;
			if (dist > distance[next.first]) continue;
			if (distance[next.first] != dist)
				usedCompany[next.first].clear();
			distance[next.first] = dist;
			if (!usedCompany[next.first].insert(next.second).second) continue;
			if (inQueue[next.first]) continue;
			inQueue[next.first] = true;
			bfs.push(next.first);
		}
	}

	if (distance.back() == inf)
		puts("-1");
	else
		printf("%d\n", distance.back());

	return 0;
}