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;

const int DMAX = 1e5+10;

struct nume{
    int node,cost;
};

vector <nume> V[5*DMAX];
priority_queue <nume> heap;
vector <int> dp(5*DMAX);
vector <bool> uz(5*DMAX);
map <pair<int,int>,int> has;

int n,m,cates;

void dijkstra();

inline bool operator<(nume x,nume y){
    return x.cost>y.cost;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    int i,x,y,z;

    cin>>n>>m;
    cates=n;
    for(i=1;i<=m;i++){
        cin>>x>>y>>z;
        if(!has[{x,z}])
           has[{x,z}]=++cates;
        if(!has[{y,z}])
           has[{y,z}]=++cates;
        V[has[{x,z}]].push_back({has[{y,z}],0});
        V[has[{y,z}]].push_back({has[{x,z}],0});
        V[has[{x,z}]].push_back({x,0});
        V[has[{y,z}]].push_back({y,0});
        V[x].push_back({has[{x,z}],1});
        V[y].push_back({has[{y,z}],1});
    }
    for(i=2;i<=cates;i++)
        dp[i]=INT_MAX/2;
    dijkstra();
    if(dp[n] == INT_MAX/2)
       cout<<"-1\n";
    else
       cout<<dp[n]<<'\n';

    return 0;
}
void dijkstra(){
    int i;
    int node;

    heap.push({1,0});
    uz[1]=true;
    while(!heap.empty()){
        node=heap.top().node;
        heap.pop();
        uz[node]=false;
        for(i=0;i<V[node].size();i++){
            if(dp[node] + V[node][i].cost < dp[V[node][i].node]){
                dp[V[node][i].node] = dp[node] + V[node][i].cost;
                if (!uz[V[node][i].node]){
                    heap.push({V[node][i].node, dp[V[node][i].node]});
                    uz[V[node][i].node] = true;
                }
            }
        }
    }
}