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: Accepted)

#include <bits/stdc++.h>

#define rep(i, n) for (int i = 0; i < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define each(x, y) for (auto &x : y)
#define all(a) (a).begin(), (a).end()
#define bit32(x) (1 << (x))
#define bit64(x) (1ll << (x))
#define sz(v) ((int) v.size())

using namespace std;

using i64 = long long;
using f80 = long double;
using vi32 = vector<int>;
using vi64 = vector<i64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << endl; exit(0); }
inline void no() { cout << "No" << endl; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); return a / gcd(a, b) * b; }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> inline void amax(T &x, T y) { x = max(x, y); }
template <typename T> inline void amin(T &x, T y) { x = min(x, y); }
template <typename T> inline T exp(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { each(x, v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }

using P = pair<int, int>;

int id = 0;
map<P, int> rid;
int getid(int u, int c) {
  if (rid.count(P(u, c))) {
    return rid[P(u, c)];
  } else {
    return rid[P(u, c)] = id++;
  }
}

void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<P>> g(n);
  rep(i, m) {
    int u, v, c;
    cin >> u >> v >> c;
    u--, v--;
    g[u].emplace_back(v, c);
    g[v].emplace_back(u, c);
    getid(u, 0);
    getid(v, 0);
    getid(u, c);
    getid(v, c);
  }
  if (!rid.count(P(0, 0)) || !rid.count(P(n - 1, 0))) {
    cout << -1 << endl;
    return;
  }
  vector<vector<P>> sg(id);
  rep(u, n) each(p, g[u]) {
    int v, c;
    tie(v, c) = p;
    sg[getid(u, c)].emplace_back(getid(v, c), 0);
    sg[getid(u, c)].emplace_back(getid(u, 0), 0);
    sg[getid(u, 0)].emplace_back(getid(v, c), 1);
  }
  vi32 dist(id, 1e9);
  deque<P> deq;
  deq.emplace_front(getid(0, 0), 0);
  while (deq.size()) {
    int u, d;
    tie(u, d) = deq.front(), deq.pop_front();
    if (dist[u] != 1e9) continue;
    dist[u] = d;
    each(p, sg[u]) {
      int v, cost;
      tie(v, cost) = p;
      if (dist[v] != 1e9) continue;
      if (cost == 0) {
        deq.emplace_front(v, d);
      } else {
        deq.emplace_back(v, d + 1);
      }
    }
  }
  if (dist[getid(n - 1, 0)] != 1e9) {
    cout << dist[getid(n - 1, 0)] << endl;
  } else {
    cout << -1 << endl;
  }
}