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<iostream>
#include<queue>
#include<unordered_set>
using namespace std;

struct edge { int to, company; };
typedef pair<int, int> P;// (minimum fare, station number)

int main()
{
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	vector<vector<edge>> es(n);
	vector<int> d(n, INT32_MAX);
	vector<unordered_set<int>> rosen(n);

	while(m--)
	{
		int pi,qi,ci;
		cin>>pi>>qi>>ci;
		pi--; qi--;
		es[pi].push_back({qi,ci});
		es[qi].push_back({pi,ci});
	}
	priority_queue<P> q;
	d[0]=0;q.push({0,0});
	while(q.size())
	{
		P p = q.top();q.pop();
		if (d[p.second] < -p.first) continue;
		for(const auto& j:es[p.second])
		{
			int dist=d[p.second];
			if(!rosen[p.second].count(j.company)) dist++;
			if (d[j.to] < dist) continue;
			if (dist < d[j.to])
			{
				d[j.to] = dist;
				rosen[j.to].clear();
				q.push({ -dist,j.to });
			}
			rosen[j.to].insert(j.company);
		}
	}
	cout<<(d[n-1]==INT32_MAX ? -1:d[n-1])<<endl;
	return 0;
}