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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;

struct UnionFind {
  int n;
  vector<int> v;
  UnionFind(int n = 0) : n(n), v(n, -1) {}
  void init(int n) {
    this->n = n;
    this->v.resize(n, -1);
  }
  bool unite(int x, int y) {
    x = find(x); y = find(y);
    if (x == y) return false;
    if (v[y] < v[x]) swap(x, y);
    v[x] += v[y]; v[y] = x;
    return true;
  }
  int find(int x) {
    return v[x] < 0 ? x : v[x] = find(v[x]);
  }
  bool same(int x, int y) {
    return find(x) == find(y);
  }
  int size(int x) {
    return -v[find(x)];
  }
};

const int INF = 1 << 30;
int N, M;
int p[200010], q[200010], c[200010];
map<pii, int> zip;
int sz;
UnionFind uf;
int hub[200010];
vector<pii> g[600010];
int dis[600010];

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  cin >> N >> M;
  for (int i = 0; i < M; i++) {
    cin >> p[i] >> q[i] >> c[i];
    p[i]--; q[i]--; c[i]--;
    if (p[i] > q[i]) swap(p[i], q[i]);
  }
  
  for (int i = 0; i < M; i++) {
    if (zip.find(pii(p[i], c[i])) == zip.end()) {
      zip[pii(p[i], c[i])] = sz++;
    }
    if (zip.find(pii(q[i], c[i])) == zip.end()) {
      zip[pii(q[i], c[i])] = sz++;
    }
  }
  uf.init(sz);
  for (int i = 0; i < M; i++) {
    uf.unite(zip[pii(p[i], c[i])], zip[pii(q[i], c[i])]);
  }

  int sz2 = 0;
  fill_n((int*)hub, 200010, -1);
  for (int i = 0; i < sz; i++) {
    int pp = uf.find(i);
    if (hub[pp] < 0) hub[pp] = sz + sz2++;
    g[i].emplace_back(hub[pp], 0);
    g[hub[pp]].emplace_back(i, 0);
  }
  sz += sz2;

  for (int i = 0; i < N; i++) {
    zip[pii(i, -1)] = sz++;
  }
  for (int i = 0; i < M; i++) {
    g[zip[pii(p[i], -1)]].emplace_back(zip[pii(p[i], c[i])], 1);
    g[zip[pii(p[i], c[i])]].emplace_back(zip[pii(p[i], -1)], 1);
    g[zip[pii(q[i], -1)]].emplace_back(zip[pii(q[i], c[i])], 1);
    g[zip[pii(q[i], c[i])]].emplace_back(zip[pii(q[i], -1)], 1);
  }

  fill_n((int*)dis, 600010, INF);
  dis[zip[pii(0, -1)]] = 0;
  priority_queue<pii, vector<pii>, greater<pii>> que;
  que.emplace(0, zip[pii(0, -1)]);
  while (!que.empty()) {
    int dd, cur;
    tie(dd, cur) = que.top(); que.pop();
    if (dd < dis[cur]) continue;
    for (pii e : g[cur]) {
      int nxt, cost;
      tie(nxt, cost) = e;
      if (dis[nxt] > dis[cur] + cost) {
        dis[nxt] = dis[cur] + cost;
        que.emplace(dis[nxt], nxt);
      }
    }
  }
  int ans = dis[zip[pii(N - 1, -1)]] / 2;
  cout << (ans == INF / 2 ? -1 : ans) << endl;

  return 0;
}