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: Memory Limit Exceeded)

/*
 * @Author: SugarSBN
 * @Date: 2018-07-14 19:06:53
 * @Language:   C++
 */
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<sstream>
#include<fstream>
using namespace std;
const int MAXN=2e7+11;
const double pi=acos(-1);
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef double dbl;
#define For(a,b,c) for (int (a)=(b);(a)<(c);(a)++)
#define foreach(iter,V) for (__typeof((V.begin()) iter=(V).begin();iter!=(V).end();iter++)
#define mp make_pair
#define fi first
#define se second
#define pb push_back
vector<pii> col[MAXN];
int col_kinds[MAXN],num_of_colkinds=0;
int fa[MAXN];
int getfa(int u){
    return fa[u] == u ? u : fa[u] = getfa(fa[u]);
}
vi new_g[MAXN];
int right_point;
int n, m;
int a, b, c;
int main(){
  //  freopen("in.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    right_point = n;
    bool vis[MAXN]={0};
    For(i,0,m){
        scanf("%d%d%d", &a, &b, &c);
        a--;
        b--;
        c--;
        if (!vis[c]){
            col_kinds[num_of_colkinds++] = c;
            vis[c] = 1;
        }
        col[c].pb(mp(a, b));
    }
    For(i, 0, n) fa[i] = i;
        int ptr[MAXN]={0};
        memset(ptr, -1, sizeof(ptr));
        For(i, 0, num_of_colkinds)
        {
            For(j, 0, col[col_kinds[i]].size())
            {
                int fx = getfa(col[col_kinds[i]][j].fi);
                int fy = getfa(col[col_kinds[i]][j].se);
                fa[fx] = fy;
            }
            For(j, 0, col[col_kinds[i]].size())
            {
                if (ptr[getfa(col[col_kinds[i]][j].fi)] < 0)
                {
                    new_g[col[col_kinds[i]][j].fi].pb(right_point);
                    new_g[right_point].pb(col[col_kinds[i]][j].fi);
                    ptr[getfa(col[col_kinds[i]][j].fi)] = right_point;
                    right_point++;
                }
                else
                {
                    new_g[col[col_kinds[i]][j].fi].pb(ptr[getfa(col[col_kinds[i]][j].fi)]);
                    new_g[ptr[getfa(col[col_kinds[i]][j].fi)]].pb(col[col_kinds[i]][j].fi);
                }
                if (ptr[getfa(col[col_kinds[i]][j].se)] < 0)
                {
                    new_g[col[col_kinds[i]][j].se].pb(right_point);
                    new_g[right_point].pb(col[col_kinds[i]][j].se);
                    ptr[getfa(col[col_kinds[i]][j].se)] = right_point;
                    right_point++;
                }
                else
                {
                    new_g[col[col_kinds[i]][j].se].pb(ptr[getfa(col[col_kinds[i]][j].se)]);
                    new_g[ptr[getfa(col[col_kinds[i]][j].se)]].pb(col[col_kinds[i]][j].se);
                }
            }
    }
    int dis[MAXN];
    For(i, 0, right_point) dis[i] = -2;
    dis[0] = 0;
    queue<int> Q;
    Q.push(0);
    while(!Q.empty()){
        int v=Q.front();
        Q.pop();
        For(i,0,new_g[v].size()){
            if(dis[new_g[v][i]]==-2){
                Q.push(new_g[v][i]);
                dis[new_g[v][i]] = dis[v] + 1;
            }
        }
    }
    printf("%d\n", dis[n - 1] / 2);
    return 0;
}