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<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iomanip>
#include<vector>
#include<queue>
#include<deque>
#include<string>
#include<utility>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<set>
#include<map>
#include<bitset>
#include<stack>
#include<ctime>
#include<unordered_map>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,int> pli;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef vector<pli> vpli;
typedef vector<vi> mat;

inline ll read(){
	ll Hashimoto=0,Kanna=1;
	char I_Love=getchar();
	while(I_Love<'0'||I_Love>'9'){
		if(I_Love=='-')Kanna=-1;
		I_Love=getchar();
	}
	while(I_Love>='0'&&I_Love<='9'){
		Hashimoto=Hashimoto*10+I_Love-'0';
		I_Love=getchar();
	}
	return Hashimoto*Kanna;
}

#define I_Love_Hashimoto_Kanna main
#define fastio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define filein(s) freopen(s,"r",stdin);
#define fileout(s) freopen(s,"w",stdout);
#define file freopen("I_Love_Hashimoto_Kanna.out","w",stdout);
#define RE cout<<"I_Love_Hashimoto_Kanna"<<endl;
#define Tone(Kanna) cout<<Kanna<<endl;
#define Tall(Hashimoto,Kanna) for(int I_Love=0;I_Love<Kanna;I_Love++)cout<<Hashimoto[I_Love]<<(I_Love==Kanna-1?"\n":" ");
#define FOR(I_Love,Hashimoto,Kanna) for(int I_Love=Hashimoto;I_Love<Kanna;I_Love++)
#define MP(Hashimoto,Kanna) make_pair(Hashimoto,Kanna)
#define REV(Kanna) reverse(Kanna.begin(),Kanna.end());
#define SORT(Kanna) sort(Kanna.begin(),Kanna.end());
#define inf (int)1e9
#define linf (ll)1e18
#define mod (inf+7)

struct val{
	int cost,u,t;
};
inline bool operator<(val a,val b){
	return a.cost>b.cost;
}
priority_queue<val>q;
int n,m;
vi col[111111];
map<int,vi>g[111111];
map<int,int>dp[111111];

int I_Love_Hashimoto_Kanna() {
  fastio;
  n=read();
  m=read();
  FOR(i,0,m){
  	int a=read()-1,b=read()-1,c=read();
  	col[a].push_back(c);
		col[b].push_back(c);
		g[a][c].push_back(b);
		g[b][c].push_back(a);
	}
	q.push({0,0,0});
	dp[0][0]=0;
	while(!q.empty()){
		int cost=q.top().cost;
		int u=q.top().u;
		int t=q.top().t;
		q.pop();
		if(dp[u][t]!=cost)continue;
		if(dp[u].find(0)==dp[u].end()||dp[u][0]>cost){
			dp[u][0]=cost;
			q.push({cost,u,0});
		}
		if(t==0){
			FOR(i,0,col[u].size()){
				int cc=col[u][i];
				if(dp[u].find(cc)==dp[u].end()||dp[u][cc]>cost+1){
					dp[u][cc]=cost+1;
					q.push({cost+1,u,cc});
				}
			}
			continue;
		}
		FOR(i,0,g[u][t].size()){
			int to=g[u][t][i];
			if(dp[to].find(t)==dp[to].end()||dp[to][t]>cost){
				dp[to][t]=cost;
				q.push({cost,to,t});
			}
		}
	}
	int ans=inf;
	for(map<int,int>::iterator itr=dp[n-1].begin();itr!=dp[n-1].end();itr++){
		ans=min(ans,itr->second);
	}
	if(ans==inf)Tone(-1)
	else Tone(ans) 
	
  
  return 0;
}