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<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<unordered_map>

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

long long int INF = 1e18;
long long int MOD = 1e9 + 7;

unordered_map<long long int, vector<long long int>> E;
unordered_map<long long int, long long int> dist;

long long int MUL = 1e9;

int main(){
  int N, M;
  cin >> N >> M;
	for(int i = 0; i < M; i++){
		long long int p, q, c;
		cin >> p >> q >> c;
		long long int pos1 = c * MUL + p, pos2 = c * MUL + q;
		E[p].push_back(pos1);
		E[pos1].push_back(p);
		E[q].push_back(pos2);
		E[pos2].push_back(q);
		E[pos1].push_back(pos2);
		E[pos2].push_back(pos1);
	}
	deque<P> que;
	que.push_back(P(1, 0));
	while(que.size() > 0){
		P p = que.front();
		que.pop_front();
		long long int pos = p.first, cost = p.second;
		if(dist.find(pos) != dist.end()){
			continue;
		}
		dist[pos] = cost;
		for(long long int to : E[pos]){
			if(pos < MUL){
				que.push_back(P(to, cost + 1));
			}else{
				que.push_front(P(to, cost));
			}
		}
	}
	if(dist.find(N) == dist.end()){
		cout << -1 << endl;
	}else{
		cout << dist[N] << endl;
	}
  return 0;
}