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)

#include <bits/stdc++.h>
#define MAXN 100005
#define tomin(a,b) (a>(b)?a=(b):a) 
#define inf 2139062143
int N,M;
struct Node{
	int p,c;
};
std::vector<Node> mp[MAXN];
std::set<int> sub[MAXN];
std::queue<int> q;
int dis[MAXN];
bool inq[MAXN];
void spfa(){
	memset(dis,127,sizeof(dis));
	q.push(1);
	dis[1]=0;inq[1]=true;
	while(!q.empty()){
		int now=q.front();
		q.pop();inq[now]=false;
		for(Node nxt:mp[now]){
			int cost=dis[now];
			if(sub[now].find(nxt.c)==sub[now].end())
				cost++;
			if(dis[nxt.p]>cost){
				dis[nxt.p]=cost;
				sub[nxt.p].clear();
				if(!inq[nxt.p]){
					q.push(nxt.p);
					inq[nxt.p]=true;
				}
			}
			if(dis[nxt.p]==cost)
				sub[nxt.p].insert(nxt.c);
		}
	}
}
int main(){
	scanf("%d%d",&N,&M);
	for(int i=1,p,q,c;i<=M;i++){
		scanf("%d%d%d",&p,&q,&c);
		mp[p].push_back((Node){q,c});
		mp[q].push_back((Node){p,c});
	}
	spfa();
	printf("%d\n",dis[N]==inf?-1:dis[N]);
	return 0;
}