Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int N=600005;
map<pair<int,int>,int>mp;
vector<pair<int,int>>e[N];
int n,m,cnt,vis[N];
priority_queue<pair<int,int>>q;
void add(int p,int q,int c,int d)
{
	e[p].push_back({q,c});
	e[q].push_back({p,d});
}
int solve(int u,int v)
{
	if(mp.count({u,v}))
		return mp[{u,v}];
	int ret=cnt++;
	add(u,ret,1,0);
	return mp[{u,v}]=ret;
}
int main()
{
	scanf("%d%d",&n,&m);
	cnt=n+1;
	for(int i=1;i<=m;i++)
	{
		int p,q,c;
		scanf("%d%d%d",&p,&q,&c);
		int a=solve(p,c),b=solve(q,c);
		add(a,b,0,0);
	}
	q.push({0,1});
	while(!q.empty())
	{
		pair<int,int> tmp=q.top();
		q.pop();
		int cost=-tmp.first,pt=tmp.second;
		if(vis[pt])
			continue;
		vis[pt]=1;
		if(pt==n)
		{
			printf("%d\n",cost);
			return 0;
		}
		for(auto it:e[pt])
			q.push({-(cost+it.second),it.first});
	}
	puts("-1");
	return 0;
}