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<vector>
#include<queue>
#include<cstring>
#include<map>
#define N 100000
#define M 200000
using namespace std;
vector<pair<int,int> > adj[2*(M+N)];
map<pair<int,int>,int> id;
int curr = 0; //# of nodes currently; id of nth node is n
int n, m;
void creati(int u, int c)
{
	int my_id = ++curr;

	adj[u].push_back({my_id,1}); //entering the hub is free, but then entering a terminal costs money
	adj[my_id].push_back({u,0}); 

	id[{u,c}] = my_id;
}
bool vis[2*(M+N)];
int d[2*(M+N)];
int main()
{
	cin >> n >> m;
	curr = n;

	while(m--)
	{
		int u, v, c; cin >> u >> v >> c;
		if(id.find({u,c})==id.end())
			creati(u,c);
		if(id.find({v,c})==id.end())
			creati(v,c);
		int a = id[{u,c}];
		int b = id[{v,c}];
		adj[a].push_back({b,0});
		adj[b].push_back({a,0});
	}
	//Start at the hub of 1, end at the hub of n

	memset(vis,false,sizeof vis);
	memset(d,-1,sizeof d);
	priority_queue<pair<int,int> > frontier;
	frontier.push({0,1});
	d[1] = 0;

	while(!frontier.empty())
	{
		int u = frontier.top().second;
		frontier.pop();
		if(!vis[u])
		{
			vis[u] = true;
			for(const auto& e : adj[u])
			{
				int v = e.first, w = e.second;
				if(d[v]==-1 || d[u]+w < d[v])
				{
					d[v] = d[u]+w;
					frontier.push({-d[v],v});
				}
			}
		}
	}
	cout<<d[n]<<endl;

	return 0;
}