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>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)   FOR(i,0,n)
#define E(x,y,c) edge{P{(x),(y)},c}
const int INF = 1e9 + 7;

using namespace std;

#define P pair<int,int>
#define MAX_V 100001
struct edge{
    P pi;
    int cost;
};

int main(){
    ios::sync_with_stdio(0),cin.tie(0);
    int v,m;
    cin >> v >> m;
    vector<map<int,vector<edge>>> G(v+1);
    REP(i,m){
        int p,q,c;
        cin >> p >> q >> c;
        G[p][c].push_back(E(q,c,0));
        G[q][c].push_back(E(p,c,0));

        G[p][c].push_back(E(p,0,0));
        G[p][0].push_back(E(p,c,1));
        G[q][c].push_back(E(q,0,0));
        G[q][0].push_back(E(q,c,1));
    }

    vector<map<int,int>> d(v+1);

    priority_queue<pair<int,P>,vector<pair<int,P>>,greater<pair<int,P>>> que;
    que.push(make_pair(0,P{1,0}));

    while(!que.empty()){
        pair<int,P> p = que.top(); que.pop();
        int vx=p.second.first,vy=p.second.second;
        //printf("[%d][%d]:%d\n",vx,vy,(int)G[vx][vy].size());
        //if(d[vx] < p.first) continue;
        REP(i,G[vx][vy].size()){
            edge e = G[vx][vy][i];
            int ex=e.pi.first,ey=e.pi.second;
            if(d[ex][ey] == 0 || d[ex][ey] > d[vx][vy] + e.cost){
                d[ex][ey] = d[vx][vy] + e.cost;
                que.push(make_pair(d[ex][ey],e.pi));
            }
        }
    }

    if(d[v][0]) cout << d[v][0] << "\n";
    else cout << "-1\n";
    return 0;
}