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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

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

typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
const int INF = int (1e9) + int (1e5);
const ll INFL = ll(2e18) + ll(1e10);
const ui MOD = 1E9 + 7;
#define FOR(i,n) for (int i=0;i<(n);++i)
#define ROF(i,x) for(int i = (x) ; i >= 0 ; --i)
#define MP make_pair
#define ALL(a) (a).begin(),(a).end()
#define ODD(x) ( ((x)&1)==0?0:1 )
#define SIGN(x) ( ((x) > 0) - ((x) < 0)  )
#define dbg(x) cerr << #x"= " << x << endl


inline ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a, ll b) { return a ? gcd(b%a, a): b; }
ll lcm(ll a, ll b) { return a / gcd(a,b) * b; }


void READ(bool _local){
    ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef _DEBUG
    if (_local)
        freopen ("in.txt", "r", stdin);
#endif
}

int n;
vector<vii> g;
int solve(){
    vector<map<int,int>> dist(n+1);
    vector<int> mnd(n+1,INF);
    mnd[1]=0;
    priority_queue<vi> q;
    q.push(vi{0,1,-1});
    while(!q.empty()){
        auto v=q.top();q.pop();
        v[0] = -v[0];
        if (v[1]==n) return v[0];
        for(ii ch:g[v[1]]){
            if (mnd[ch.first] < v[0]) continue;
            int d = v[0] + (v[2]!=ch.second);
            auto it=dist[ch.first].find(ch.second);
            if (it==dist[ch.first].end()){
                dist[ch.first][ch.second] = d;
                q.push(vi{-d,ch.first,ch.second});
                mnd[ch.first] = min(mnd[ch.first],d);
            }else if (it->second>d){
                it->second=d;
                q.push(vi{-d,ch.first,ch.second});
                mnd[ch.first] = min(mnd[ch.first],d);
            }
        }
    }
    return -1;
}

int main() {
    READ(0);
    int m;cin>>n>>m;
    g.resize(n+1);
    FOR(i,m){
        int u,v,c;cin>>u>>v>>c;
        g[u].emplace_back(v,c);
        g[v].emplace_back(u,c);
    }
    cout << solve();
    return 0;
}