Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=int(a);i<int(b);++i)
#define SIZE 100005
#define INF 1000000005LL
#define MOD 1000000007

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

int N,M;

int V;
struct edge{int to,cost;};
vector<edge> G[SIZE*5];
int d[SIZE*5];
map<int,int> S[SIZE];
void dijkstra(int s);
int t;

void add_edge(int p,int q,int c){
	if(S[p].find(c) == S[p].end()){
		S[p][c] = ++t;
		G[p].push_back((edge){t,0});
		G[t].push_back((edge){p,1});
	}
	if(S[q].find(c) == S[q].end()){
		S[q][c] = ++t;
		G[q].push_back((edge){t,0});
		G[t].push_back((edge){q,1});
	}
	int t1 = S[p][c];
	int t2 = S[q][c];
	G[t1].push_back((edge){t2,0});
	G[t2].push_back((edge){t1,0});
}


int main()
{
	cin >> N >> M;
	t = N;
	rep(i,0,M){
		int p,q,c;
		cin >> p >> q >> c;
		add_edge(p,q,c);
	}
	V = t+1;
	dijkstra(1);
	if(d[N]>=INF)d[N]=-1;
	cout << d[N] << endl;

	return 0;
}

void dijkstra(int s){
	priority_queue<P, vector<P>, greater<P>> que;
	fill(d,d+V,INF);
	d[s] = 0;
	que.push(P(0,s));

	while(!que.empty()){
		P p = que.top();que.pop();
		int v = p.second;
		if(d[v] < p.first)continue;
		for(int i=0;i<G[v].size();i++){
			edge e = G[v][i];
			if(d[e.to] > d[v] + e.cost){
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to],e.to));
			}
		}
	}
}