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>
using namespace std;
typedef pair<int, int> ii;

struct DisjointSet{
	vector<int> p;
	void init(int n){
		p.resize(n); iota(p.begin(), p.end(), 0);
	}
	int root(int i){
		return p[i] == i ? i : p[i] = root(p[i]);
	}
	void unify(int i, int j){
		int x = root(i), y = root(j);
		if(x==y) return;
		p[x]=y;
	}
} dsu;

vector<vector<ii> > g;
vector<vector<int> > G;
int n, m;

int bfs(){
	queue<int> q;
	vector<int> dist(n+m, -2);
 dist[0] = 0; q.push(0);
	while(!q.empty()){
		int u = q.front(); q.pop();
		for(int v: G[u]){
			if(dist[v] == -2 || dist[v] > dist[u]+1){
				dist[v] = dist[u]+1;
				q.push(v);
			}
		}
	}
	return dist[n-1];
}

signed main(){
	cin >> n >> m;
	g.assign(n, vector<ii>());
	for(int i=0; i< m; ++i){
		int u, v, c; cin >> u >> v >> c;
		g[--u].push_back({c, i});
		g[--v].push_back({c, i});
	}
	dsu.init(m);
	G.assign(n+m, vector<int>());
	for(int i=0; i< n; ++i){
		sort(g[i].begin(), g[i].end());
		//unify same node-edges
		for(int j=1; j< g[i].size(); ++j){
			if(g[i][j].first == g[i][j-1].first){
				dsu.unify(g[i][j].second, g[i][j-1].second);
			}
		}
	}
	//add connection from node-node to node-edge
	for(int i=0; i< n; ++i){
		for(ii v : g[i]){
			int r = n + dsu.root(v.second);
			G[r].push_back(i);
			G[i].push_back(r);
		}
	}
	cout << (bfs()/2);
}