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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

typedef pair<int,int> P;

int nextInt() {
  int x; scanf("%d", &x); return x;
}

const int MAX_N = 100000 + 10;

vector<P> es[MAX_N];

struct State {
  int pos;
  int pc;
  int cost; 
  State(int a, int b, int c) : pos(a), pc(b), cost(c) {}
};

int main() {
  int N = nextInt();
  int M = nextInt();
  for (int i = 0; i < M; i++) {
    int p = nextInt() - 1;
    int q = nextInt() - 1;
    int c = nextInt();
    es[p].emplace_back(q, c);
    es[q].emplace_back(p, c);
  }

  int ans = -1;
  map<P, int> memo; // 0なら存在しないノード、最小値は 1 
  deque<State> pq;
  pq.emplace_back(0, 0, 1); // 今いる駅, 最後に使った会社, コスト
  memo[P(0,0)] = 1;
  while (!pq.empty()) {
    const State state = pq.front();
    pq.pop_front();
    // const int cost = memo[P(state.pos, state.pc)];
    // if (state.cost != cost) continue;
    const int cost = state.cost;
    if (state.pos == N - 1) {
      ans = cost - 1;
      break;
    }
    for (auto e : es[state.pos]) {
      int next_cost = cost + (state.pc == e.second ? 0 : 1);
      auto f = memo.find(e);
      if (f == memo.end() || f->second > next_cost) {
        memo[e] = next_cost;
        if (state.pc == e.second) 
          pq.emplace_front(e.first, e.second, next_cost);
        else
          pq.emplace_back(e.first, e.second, next_cost);
      }
    }
  }
  cout << ans << endl;
  return 0;
}