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>
#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;
using namespace std;
 
typedef long long ll;

const int N = 1e5 + 7;
const int inf = 1e9 + 7;

vector <pair <int, int> > g[N];

const int RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
struct chash {
    int operator()(int x) const { return x ^ RANDOM; }
};
gp_hash_table<int, int, chash> d[N];
gp_hash_table<int, int, chash> mn;

bool used[N];

struct triple {
    int v, c, d;
};

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        a--, b--;
        g[a].push_back({b, c});
        g[b].push_back({a, c});
        d[a][c] = inf;
        d[b][c] = inf;
        mn[a] = inf;
        mn[b] = inf;
    }
    queue <triple> q, go;
    d[0][0] = 0;
    mn[0] = 0;
    q.push({0, 0, 0});
    while (!q.empty() || !go.empty()) {
        if (q.empty()) {
            swap(q, go);
        }
        auto cur = q.front();
        int v = cur.v;
        int cur_c = cur.c;
        int cur_d = cur.d;
        q.pop();
        if (cur_d > d[v][cur_c] || cur_d > mn[v] || used[v]) {
            continue;
        }
        used[v] = true;
        for (auto to : g[v]) {
            int u = to.first;
            int go_c = to.second;
            if (d[u][go_c] > d[v][cur_c] + (cur_c != go_c)) {
                d[u][go_c] = d[v][cur_c] + (cur_c != go_c);
                int f = (cur_c != go_c);
                if (f) {
                    go.push({u, go_c, d[u][go_c]});
                } else {
                    q.push({u, go_c, d[u][go_c]});    
                }
                mn[u] = min(mn[u], d[u][go_c]);
            }
        }
    }
    int ans = inf;
    for (auto to : d[n - 1]) {
        ans = min(ans, to.second);
    }
    cout << (ans == inf ? -1 : ans) << '\n';
    return 0;
}