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>
using namespace std;

const int INF = 0x3f3f3f3f;

struct Node{
	int to, dis, p;
	Node(int _to, int _dis){to = _to, dis = _dis;}
	Node(int _to, int _dis, int _p){to = _to, dis = _dis, p = _p;}
	bool operator < (const Node &A) const{
		return dis > A.dis;
	}
};

vector <Node> from[100010];
int n, m;
int dis[100010][2];
int P[100010][2];
priority_queue <Node> q;

void Dij(int s, int p){
	for (int i = 1; i <= n; i++){
		dis[i][p] = INF;
	}
	dis[s][p] = 1;
	q.push(Node(s, 1, -1));
	while (!q.empty()){
		Node t = q.top(); q.pop();
		int u = t.to;
		if (t.dis > dis[u][p]) continue;
		for (int i = 0; i < from[u].size(); i++){
			Node tmp = from[u][i];
			int v = tmp.to;
			int new_c = t.p == -1 ? 0 : tmp.dis != t.p;
			if (dis[v][p] > dis[u][p] + new_c){
				dis[v][p] = dis[u][p] + new_c;
				P[v][p] = tmp.dis;
				q.push(Node(v, dis[v][p], tmp.dis));
			}
		}
	}
}

int main(){
	scanf("%d%d", &n, &m);
	for (int i = 0; i < m; i++){
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		from[u].push_back(Node(v, c));
		from[v].push_back(Node(u, c));
	}
	Dij(1, 0);
	Dij(n, 1);
	int ans = min(dis[n][0], dis[1][0]);
	for (int i = 1; i <= n; i++){
		ans = min(ans, dis[i][0] + dis[i][1] - 1 + (P[i][0] != P[i][1]));
	}
	printf("%d\n", ans == INF ? -1 : ans);
	return 0;
}