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 N = 200005;

struct Edge {
  int v, c;
  Edge(const int &v = 0, const int &c = 0) : v(v), c(c) {}
};

vector<Edge> g[N];
int d[N], vis[N];

queue<int> q;

void dfs(int u, int C) {
  q.emplace(u);
  vis[u] = 1;
  for (auto &e:g[u])
    if (!vis[e.v] && e.c == C) {
      d[e.v] = d[u];
      dfs(e.v, C);
    }
}

int dijkstra(int s, int t) {
  memset(d, 0xff, sizeof d);
  d[s] = 0;
  vis[s] = 1;
  q.emplace(s);
  while (q.size()) {
    auto u = q.front();
    q.pop();

    for (auto &e : g[u])
      if (!vis[e.v]) {
        d[e.v] = d[u] + 1;
        dfs(e.v, e.c);
      }
  }
}

int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    g[u].emplace_back(v, c);
    g[v].emplace_back(u, c);
  }
  dijkstra(1, n);
  cout << d[n] << endl;
  return 0;
}