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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

typedef pair<int,int> pii;

int N,M;
vector<pii> E[1000010];
int unit = 0;
int mem[200010];
int added[200010];
queue<int> Q;
vector<int> edge[200010];
vector<int> line[200010];
priority_queue<pii, vector<pii> ,greater<pii> > PQ;
int D[100010];
vector<int> V[100010];
vector<int> including;
int L;
bool visited[200010] = {0};
vector<int> C;
int i;

int main()
{
    scanf("%d%d",&N,&M);
    for(int i = 0; i <= N; i++)
    {
        mem[i] = added[i] = D[i] = -1;
    }
    for(int i = 0; i < M; i++)
    {
        int p,q,c;
        scanf("%d%d%d",&p,&q,&c);
        E[c].push_back(make_pair(p,q));
        C.push_back(c);
    }
    for(int l = 0; l < C.size(); l++)
    {
        i = C[l];
        if(!E[i].empty())
        {
            for(int j = 0; j < E[i].size(); j++)
            {
                if(added[E[i][j].first] != i)
                {
                    including.push_back(E[i][j].first);
                    added[E[i][j].first] = i;
                }
                if(added[E[i][j].second] != i)
                {
                    including.push_back(E[i][j].second);
                    added[E[i][j].second] = i;
                }
                V[E[i][j].first].push_back(E[i][j].second);
                V[E[i][j].second].push_back(E[i][j].first);
            }
            L = unit;
            for(int j = 0; j < including.size(); j++)
            {
                if(mem[including[j]] < L)
                {
                    Q.push(including[j]);
                    while(!Q.empty())
                    {
                        int now = Q.front();
                        Q.pop();
                        if(mem[now] == unit)
                        {
                            continue;
                        }
                        mem[now] = unit;
                        edge[now].push_back(unit);
                        line[unit].push_back(now);
                        for(int k = 0; k < V[now].size(); k++)
                        {
                            Q.push(V[now][k]);
                        }
                    }
                    unit++;
                }
            }
            for(int j = 0; j < including.size(); j++)
            {
                while(V[including[j]].size())
                {
                    V[including[j]].pop_back();
                }
            }
            while(including.size())
            {
                including.pop_back();
            }
        }
    }
    PQ.push(make_pair(0,1));
    while(PQ.size())
    {
        pii now = PQ.top();
        PQ.pop();
        if(D[now.second] != -1)
        {
            continue;
        }
        D[now.second] = now.first;
        for(int i = 0; i < edge[now.second].size(); i++)
        {
            if(visited[edge[now.second][i]])
            {
                continue;
            }
            visited[edge[now.second][i]] = true;
            for(int j = 0; j < line[edge[now.second][i]].size(); j++)
            {
                PQ.push(make_pair(now.first + 1,line[edge[now.second][i]][j]));
            }
        }
    }
    printf("%d\n",D[N]);
    return 0;
}