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> // {{{

#define ARG4(_1, _2, _3, _4, ...) _4
#define rep(...) ARG4(__VA_ARGS__, FOR, REP)(__VA_ARGS__)
#define REP(i, a) FOR(i, 0, a)
#define FOR(i, a, b) for (int i = (a); i < (int)(b); ++i)
#define rrep(...) ARG4(__VA_ARGS__, RFOR, RREP)(__VA_ARGS__)
#define RREP(i, a) RFOR(i, 0, a)
#define RFOR(i, a, b) for (int i = (b)-1; i >= (int)(a); --i)
#define ALL(c) (c).begin(), (c).end()
#define TEN(n) ((ll)(1e##n))
#define pb emplace_back
#define mp make_pair
#define fst first
#define snd second
#define let const auto
#define USE1(T) template <typename T>
#define USE2(T, U) template <typename T, typename U>
#define I_ inline
#define mygc(c) (c) = getchar_unlocked()
#define mypc(c) putchar_unlocked(c)

USE1(T) using duo = std::pair<T, T>;
USE1(T) using vec = std::vector<T>;
using ll = long long;
// clang-format off
USE2(T,U) I_ bool chmax(T&x,U a){return x<a&&(x=a,1);}
USE2(T,U) I_ bool chmin(T&x,U a){return a<x&&(x=a,1);}
USE1(T=int) I_ T in(){T x;std::cin>>x;return x;}
USE1(T=int) I_ vec<T> in(int n){vec<T> v;v.reserve(n);rep(i,n)v.pb(in<T>());return v;}
USE1(T) vec<T> ndvec(T v,int n){return vec<T>(n, v);}
USE2(T,...Ts) auto ndvec(T v,int n,Ts...ns)->vec<decltype(ndvec(v,ns...))>{return ndvec(ndvec(v,ns...),n);}
USE1(T) I_ void pr(T x){std::cout<<x<<'\n';}
USE2(T,...Ts) I_ void pr(T x,Ts...xs){std::cout<<x<<' ';pr(xs...);}
I_ int rd(){int x=0,m=0,k;for(;;){mygc(k);if(k=='-'){m=1;break;}if('0'<=k&&k<='9'){x=k-'0';break;}}for(;;){mygc(k);if(k<'0'||'9'<k)break;x=x*10+k-'0';}return x;}
I_ void wr(int x,char c='\n'){int s=0,m=0;char b[32];if(x<0)m=1,x=-x;for(;x;x/=10)b[s++]=x%10;if(!s)b[s++]=0;if(m)mypc('-');for(;s--;)mypc(b[s]+'0');mypc(c);}
struct IoSetup {
  IoSetup(){
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);
    std::cout.precision(10);
    std::cerr.precision(10);
  }
} iosetup;
// clang-format on
// }}}
using namespace std;
const int inf = 1001001001;
const ll infl = 1001001001001001001ll;
const int dd[] = {0, 1, 0, -1, 0};

struct Edge {
  int to;
  int com;
  int cost;
  Edge(int to, int com, int cost = 1) : to(to), com(com), cost(cost) {}
};
using Graph = vec<vec<Edge>>;

int bfs01(const Graph& g, int src, int dst) {
  static unordered_map<ll, int> cost;
  cost.clear();
  deque<Edge> Q;
  Q.pb(src, 0, 0);

  int ans = inf;
  while (not Q.empty()) {
    auto e = Q.front();
    Q.pop_front();
    for (const auto& u : g[e.to]) {
      ll ehash = u.to * 1001001ll + u.com;
      int min_cost = inf;
      if (cost.count(ehash)) { min_cost = cost[ehash]; }
      int nxt_cost = e.cost + (u.com != e.com);
      if (min_cost <= nxt_cost) continue;
      cost[ehash] = nxt_cost;
      if (u.com == dst) chmin(ans, nxt_cost);
      if (u.com == e.com) {
        Q.emplace_front(u.to, u.com, nxt_cost);
      } else {
        Q.pb(u.to, u.com, nxt_cost);
      }
    }
  }
  return ans;
}

signed main() {
  int N = in(), M = in();
  Graph G(N + 1);
  rep(i, M) {
    int p = in(), q = in(), c = in();
    G[p].pb(q, c);
    G[q].pb(p, c);
  }
  int ans = min(bfs01(G, 1, N), bfs01(G, N, 1));
  if (ans == inf) {
    pr(-1);
  } else {
    pr(ans);
  }
  return 0;
}