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> pi;
typedef pair<int, pi> pii;


map<pi, int> ma;
vector<pi> adj[555555];
deque<int> dq;

int n, m, p, q, c, cur, dist[555555];
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
	cin >> n >> m;
	cur = n+1;
	for(int i = 0; i < m; i++){
		cin >> p >> q >> c;
		if(ma[pi(p, c)] == 0) ma[pi(p, c)] = cur++;
		if(ma[pi(q, c)] == 0) ma[pi(q, c)] = cur++;
		adj[p].push_back(pi(1, ma[pi(p, c)]));
		adj[ma[pi(p, c)]].push_back(pi(0, p));
		adj[q].push_back(pi(1, ma[pi(q, c)]));
		adj[ma[pi(q, c)]].push_back(pi(0, q));
		adj[ma[pi(p, c)]].push_back(pi(0, ma[pi(q, c)]));
		adj[ma[pi(q, c)]].push_back(pi(0, ma[pi(p, c)]));
	}
	memset(dist, -1, sizeof(dist));
	dist[1] = 0;
	dq.push_back(1);
	while(!dq.empty()){
		int f = dq.front(); dq.pop_front();
		for(auto it:adj[f]){
			if(dist[it.second] == -1 || dist[it.second] > dist[f]+it.first){
				dist[it.second] = dist[f]+it.first;
				if(it.first) dq.push_back(it.second);
				else dq.push_front(it.second);
			}
		}
	}
	cout << dist[n];
}