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 <iostream>
#include <algorithm>
#include <vector>
#include <iomanip>
#include <map>
#include <queue>

static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using namespace std;

template<class T>
constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;

template <typename T>
using GPQ = priority_queue<T, vector<T>, greater<T>>;
using P = pair<int, int>;
int main() {
    int n, m;
    cin >> n >> m;
    vector<int> dis(n, INF<int>);
    vector<map<int, int>> com(n);
    vector<vector<P>> G(n);
    for (int i = 0; i < m; ++i) {
        int p, q, c;
        scanf("%d %d %d", &p, &q, &c);
        G[p-1].emplace_back(q-1, c);
        G[q-1].emplace_back(p-1, c);
    }
    dis[0] = 0;
    GPQ<P> Q;
    Q.emplace(0, 0);
    while (!Q.empty()){
        int cost;
        int i;
        tie(cost, i) = Q.top(); Q.pop();
        if(dis[i] < cost) continue;
        for (auto &&e : G[i]) {
            int cost2 = cost + (com[i][e.second] == 0);
            if(dis[e.first] < cost2) continue;
            if(dis[e.first] > cost2) {
                com[e.first].clear();
                Q.emplace(cost2, e.first);
            }
            dis[e.first] = cost2;
            com[e.first][e.second]++;
        }
    }
    cout << (dis[n-1] == INF<int> ? -1 : dis[n-1]) << "\n";
     return 0;
}