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>
using namespace std;

class Graph {
  private:
    int n;
    vector<vector<pair<int, int>>> ad;

  public:
    Graph(int n) :
        n(n), ad(n + 1) { }

    void addEdge(int x, int y, int z) {
        static map<pair<int, int>, int> comp;
        if (!comp[{x, z}]) { comp[{x, z}] = ++n; ad.emplace_back(); }
        if (!comp[{y, z}]) { comp[{y, z}] = ++n; ad.emplace_back(); }

        ad[comp[{x, z}]].emplace_back(comp[{y, z}], 0);
        ad[comp[{y, z}]].emplace_back(comp[{x, z}], 0);

        ad[x].emplace_back(comp[{x, z}], 0);
        ad[y].emplace_back(comp[{y, z}], 0);

        ad[comp[{x, z}]].emplace_back(x, 1);
        ad[comp[{y, z}]].emplace_back(y, 1);
    }

    vector<int> dijkstra(int src) {
        vector<int> dp(n + 1, 1e9);
        dp[src] = 0;

        priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
        pq.emplace(0, src);

        vector<bool> vis(n + 1);
        while (!pq.empty()) {
            int node = pq.top().second;
            pq.pop();

            if (vis[node])
                continue;
            vis[node] = true;

            for (auto arc : ad[node])
                if (dp[node] + arc.second < dp[arc.first]) {
                    dp[arc.first] = dp[node] + arc.second;
                    pq.emplace(dp[arc.first], arc.first);
                }
        }
        return dp;
    }
};

int main() {
    int n, m;
    cin >> n >> m;

    Graph graph(n);
    for (int i = 0; i < m; i++) {
        int x, y, z; cin >> x >> y >> z;
        graph.addEdge(x, y, z);
    }

    int sol = graph.dijkstra(1)[n];
    cout << (sol == 1e9 ? -1 : sol) << '\n';
    return 0;
}