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: Wrong Answer)

#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  for(int i=0;i<(n);++i)
#define REPr(i,n) for(int i=(n)-1;i>=0; --i)
#define FORq(i, m, n) for(int i = (m);i <= (n);++i)
#define FORqr(i, m , n) for(int i = (n);i >=(m);--i)
#define SCD(n) scanf("%d",&n)
#define SCD2(m,n) scanf("%d%d",&m,&n)
#define SCD3(m,n,k) scanf("%d%d%d",&m,&n,&k)
#define SCLLD(n) scanf("%lld",&n)
#define SCLLD2(m,n) scanf("%lld%lld",&m,&n)
#define SCLLD3(m,n,k) scanf("%lld%lld%lld",&m,&n,&k)
#define PB push_back
#define MP make_pair
#define ARSCD(A,N) REP(i,N){SCD(A[i]);}
#define ARSCD1(A,N) FORq(i,1,N){SCD(A[i]);}
#define VSCD(v,N) REP(i,N){int x; SCD(x); v.PB(x);}
#define VSCLLD(v,N) REP(i,N){long long x; SCLLD(x); v.PB(x);}
#define PRINTD(n) printf("%d\n",n)
#define PRINTLLD(n) printf("%lld\n",n)
#define DEBUG printf("%s\n","debug")
#define fst first
#define snd second
#define SIN(x,S) (S.count(x) != 0)
using namespace std;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector < VI > VVI;
typedef vector<long long> VL;
typedef long long ll;
typedef long long integer;
////////////////////////////////////////////////////////////////////
struct Vertex {
	// edge information
	vector<int> edges_to; //to vertex number
	vector<ll> edges_cost; // edge cost
	
	//for dijkstra
	bool done; //conform or not
	ll cost = -2; // now cost
	int from;
	
	bool operator<(const Vertex& another) const { // priority --> min
        return another.cost < cost; // order(cost)
    }
	
	bool operator==(const Vertex& another) const {
        return another.cost == cost; // order(cost)
    }
};
Vertex V[1500002];

int main(){
    
    const int K = 200001;

    int N,M;
    SCD2(N,M);

    REP(i,M){
        int p,q,c;
        SCD3(p,q,c);

        V[K+c].edges_to.PB(p);
        V[K+c].edges_cost.PB(1);
        V[p].edges_to.PB(c+K);
        V[p].edges_cost.PB(1);

        V[K+c].edges_to.PB(q);
        V[K+c].edges_cost.PB(1);
        V[q].edges_to.PB(c+K);
        V[q].edges_cost.PB(1);

    }

    priority_queue<int> Q;

    V[1].cost = 0; //start
	Q.push(1);
	
	while (!Q.empty()){
		Vertex nowV = V[Q.top()]; Q.pop();
		nowV.done = true;
		
		REP(i,nowV.edges_to.size()){
			int nto = nowV.edges_to[i];
			ll ncost = nowV.cost + nowV.edges_cost[i];
			if (((V[nto].cost) < 0) or (ncost < V[nto].cost)){  //not visited or required refresh
				V[nto].cost = ncost;
				Q.push(nto);
			
			}
			
		}
	}

    PRINTD(V[N].cost / 2);

}