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 <vector>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <iomanip>
#include <cassert>
using namespace std;

#define GET_ARG(a,b,c,F,...) F
#define REP3(i,s,e) for (i = s; i <= e; i++)
#define REP2(i,n) REP3 (i,0,(int)(n)-1)
#define REP(...) GET_ARG (__VA_ARGS__,REP3,REP2) (__VA_ARGS__)
#define RREP3(i,s,e) for (i = s; i >= e; i--)
#define RREP2(i,n) RREP3 (i,(int)(n)-1,0)
#define RREP(...) GET_ARG (__VA_ARGS__,RREP3,RREP2) (__VA_ARGS__)
#define DEBUG(x) cerr << #x ": " << x << endl

constexpr int INF = 1e8;

vector<pair<int,int>> e[100000]; // (c,q)
int dist[100000];

int main(void) {
    int i, n, m;
    scanf("%d%d",&n,&m);
    REP (i,m) {
        int p, q, c;
        scanf("%d%d%d",&p,&q,&c);
        p--; q--;
        e[p].emplace_back(c,q);
        e[q].emplace_back(c,p);
    }
    priority_queue<pair<int,pair<int,int>>> q; // (-d,(c,v))
    q.push(make_pair(0,make_pair(-1,0)));
    REP (i,n) dist[i] = INF;
    dist[0] = 0;
    while (!q.empty()) {
        int c = q.top().second.first;
        int v = q.top().second.second;
        q.pop();
        for (auto x: e[v]) {
            if (dist[x.second] > dist[v] + (c != x.first)) {
                dist[x.second] = dist[v] + (c != x.first);
                q.push(make_pair(-dist[x.second],x));
            }
        }
    }
    if (dist[n-1] == INF)
        printf("-1\n");
    else
        printf("%d\n",dist[n-1]);
    return 0;
}