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 <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <random>
#include <set>
#include <string>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<double, int> pdi;
typedef pair<int, pii> pipii;
typedef vector<vector<ll>> matrix;

map<pii, vector<int>> edgesOut; // <vertex, color> -> out
vector<pii> edges[100005]; // <vertex, color>
bool seen[100005];
map<pii, int> dp;

void solve() {
  int n, m;
  cin >> n >> m;
  priority_queue<pipii> q; // <distance, <vertex, color>>
  while(m--) {
    int a, b, c;
    cin >> a >> b >> c;
    edges[a].push_back({b, c});
    edges[b].push_back({a, c});
    edgesOut[{a, c}].push_back(b);
    edgesOut[{b, c}].push_back(a);
    if(!dp.count({1, c})) {
      dp[{1, c}] = 1;
      q.push({-1, {1, c}});
    }
  }
  int ret = -1;
  while(!q.empty()) {
    pipii curr = q.top();
    q.pop();
    if(curr.second.first == n) {
      ret = -curr.first;
      break;
    }
    if(dp[curr.second] != -curr.first) continue;
    if(!seen[curr.second.first]) {
      seen[curr.second.first] = true;
      for(pii out: edges[curr.second.first]) {
        pii nextKey = out;
        if(!dp.count(nextKey) || dp[nextKey] > 1 - curr.first) {
          dp[nextKey] = 1 - curr.first;
          q.push({-dp[nextKey], nextKey});
        }
      }
    }
    if(edgesOut.count(curr.second)) {
      for(int out: edgesOut[curr.second]) {
        pii nextKey = {out, curr.second.second};
        if(!dp.count(nextKey) || dp[nextKey] > -curr.first) {
          dp[nextKey] = -curr.first;
          q.push({-dp[nextKey], nextKey});
        }
      }
    }
  }
  cout << ret << "\n";
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);
  solve();
}