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: Wrong Answer)

//hi
#include<bits/stdc++.h>
using namespace std;
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define inf 100000000000000LL
typedef long long int LL;
struct hashfunc {
	template<typename T, typename U>
	size_t operator()(const pair<T, U> &x) const {
		return hash<T>()(x.F) ^ hash<U>()(x.S);
	}
};
vector< pair<int,int> > v[100004];
unordered_map< pair<int,int>, LL ,hashfunc> d;  //pos, last lane
unordered_set< pair<int,int> ,hashfunc> vis; 
priority_queue< pair<LL, pair<int,int> > > pq; //pos, last lane
LL best[100004];
int main(void){
    int n,m;
    scanf("%d%d",&n,&m);
    int i;
    for(i=0;i<m;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		v[a].PB(MP(b,c));
		v[b].PB(MP(a,c));
	}
	for(i=1;i<=n;i++) best[i]=inf;
	for(i=0;i<v[1].size();i++){
		d[MP(1,v[1][i].S)]=1;
		pq.push(MP(-1,MP(1,v[1][i].S)));
	}
	while(!pq.empty()){
		pair<LL, pair<int,int> > now;
		int ok=0;
		while(!pq.empty()){
			//printf(" top %d %d\n",pq.top().S.F, pq.top().S.S);
			if(vis.find(pq.top().S) != vis.end()){
				pq.pop();
			}else if(-pq.top().F>best[pq.top().S.F]+1){
				vis.insert(pq.top().S);
				pq.pop();
			}else{
				now=pq.top();
				pq.pop();
				ok=1;
				break;
			}
		}
		if(!ok) break;
		int pt=now.S.F;
		if(pt==n) break;
		vis.insert(now.S);
		best[pt]=min(best[pt],-now.F);
		//printf("%d %d : %lld\n",now.S.F, now.S.S, now.F);
		for(i=0;i<v[pt].size();i++){
			int npt=v[pt][i].F;
			pair<int,int> nxt=MP(npt,v[pt][i].S);
			int g=(v[pt][i].S == now.S.S? 0:1) + (-now.F);
			if(d.find(nxt) == d.end()){
				d[nxt] = g;
				//printf("  %d %d : %d\n",nxt.F,nxt.S,g);
				pq.push(MP(-g, nxt)); 
			}else{
				if(vis.find(nxt) != vis.end()) continue;
				if(d[nxt] > g){
					d[nxt]=g;
				//	printf("  %d %d : %d\n",nxt.F,nxt.S,g);
					pq.push(MP(-g, nxt)); 
				}
			}
		}
	}
	LL ans=inf;
	for(i=1;i<=m;i++){
		if(d.find(MP(n,i)) == d.end()) continue;
		ans=min(ans, d[MP(n,i)]);
	}
	if(ans!=inf) printf("%lld\n",ans);
	else printf("-1\n");
    return 0;
}