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 <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#define INF 1e+9
#define mp make_pair
using namespace std;

struct edge{ int to,comp,cost; };
typedef pair<int,int> VC;//vertex,company
typedef pair<int,VC> P;//cost

int main(){
	int n,m;
	map<VC,int> d;
	set<int> cs[100000];
	map<VC,vector<edge>> G;
	cin >> n >> m;
	for(int i = 0;i < m;i++){
		int a,b,c;
		cin >> a >> b >> c; a--;b--;c--;
		G[mp(a,c)].push_back({b,c,0});
		G[mp(b,c)].push_back({a,c,0});
		cs[a].insert(c);
		cs[b].insert(c);
	}
	for(int i = 0;i < n;i++){
		for(auto it = cs[i].begin();it != cs[i].end();it++){
			d[mp(i,*it)] = INF;
			G[mp(i,*it)].push_back({i,-1,0});
			G[mp(i,-1)].push_back({i,*it,1});
		}
		d[mp(i,-1)] = INF;
	}
	d[mp(0,-1)] = 0;
	priority_queue<P,vector<P>,greater<P> > que;
	que.push(mp(0,mp(0,-1)));
	while(!que.empty()){
		P p = que.top();que.pop();
		int cost = p.first,v = p.second.first,comp = p.second.second;
		if(d[mp(v,comp)] < cost) continue;
		for(int i = 0;i < G[mp(v,comp)].size();i++){
			edge e = G[mp(v,comp)][i];
			if(d[mp(e.to,e.comp)] > d[mp(v,comp)] + e.cost){
				d[mp(e.to,e.comp)] = d[mp(v,comp)] + e.cost;
				que.push(mp(d[mp(e.to,e.comp)],mp(e.to,e.comp)));
			}
		}
	}
	if(d[mp(n - 1,-1)] == INF) cout << -1 << endl;
	else cout << d[mp(n - 1,-1)] << endl;
	return 0;
}