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<iostream>
#include<queue>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
#define pb push_back
int dy[]={0, 0, 1, -1, 1, 1, -1, -1};
int dx[]={1, -1, 0, 0, 1, -1, -1, 1};

#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define rep(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
#define rrep(i,n) RREP(i,n)
#define mp make_pair
#define fi first
#define sc second
#define INF 1e12

ll n,m;

struct edge {ll to,com;};

vector<edge> G[200001];

queue<ll> que;
ll d[200001];

void dfs(ll v,ll dd,ll cm) {
	d[v] = dd;
	que.push(v);
	
	rep(i,G[v].size()) {
		if(d[G[v][i].to]== INF && G[v][i].com == cm){
			dfs(G[v][i].to,dd,cm);
		}
	}
}
int main(){
	cin >> n >> m;

	rep(i,m) {
		ll p,q,c;
		cin >> p >> q >> c;
		p--;q--;
		G[p].pb((edge){q,c});
		G[q].pb((edge){p,c});
 	}
	rep(i,n)d[i] = INF;

	d[0] = 0;
	que.push(0);

	while(!que.empty()) {
		ll v = que.front();
		que.pop();
		rep(i,G[v].size()) {
			if(d[G[v][i].to]==INF)
				dfs(G[v][i].to,d[v]+1,G[v][i].com);
		}
	}
	if(d[n-1]==INF)cout << -1 << endl;
	else cout << d[n-1] << endl;

	return 0;
}