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>
    #define INF 4000001
    using namespace std;
    int n,m;
    struct edge{int to,co;};
    vector<edge> g[100010];
    int dis[100010];
    set<int> disc[100010];
    int main()
    {
    	ios::sync_with_stdio(0);
    	cin.tie(0);
     
    	cin>>n>>m;
    	for(int i=0;i<m;i++)
    	{
    		int a,b,c;
    		cin>>a>>b>>c;
    		g[a].push_back({b,c});
    		g[b].push_back({a,c});
    	}
     
    	dis[1]=0;
    	for(int i=2;i<=n;i++)
    		dis[i]=INF;
    	priority_queue<pair<int,int>> q;
    	q.push({-dis[1],1});
    	while(q.size())
    	{
    		int now=q.top().second;
    		int d=-q.top().first;
    		q.pop();
    		if(dis[now]<d) continue;
    		for(edge j:g[now])
    		{
    			//cout<<j.to<<endl;
    			int nd=dis[now];
    			if(!disc[now].count(j.co)) nd++;
    			if(nd<dis[j.to])
    			{
    				dis[j.to]=nd;
    				disc[j.to].clear();
    				disc[j.to].insert(j.co);
    				q.push({-dis[j.to],j.to});
    			}
    			else if(nd==dis[j.to])
    			{
    				disc[j.to].insert(j.co);
    			}
    		}
    	}
    	if(dis[n]==INF)
    		cout<<-1<<endl;
    	else
    		cout<<dis[n]<<endl;
    }