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;
 
using P = pair<int, int>;
using PI = pair<P, int>;
using Graph = map<P, vector<PI> >;
 
int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n, m;
	cin >> n >> m;
	Graph g;
	for (int i = 0; i < m; i++) {
		int p, q, c;
		cin >> p >> q >> c;
		P x0(p, 0), x1(p, c), y0(q, 0), y1(q, c);
		g[x1].emplace_back(y1, 0);
		g[y1].emplace_back(x1, 0);
		g[x1].emplace_back(x0, 0);
		g[x0].emplace_back(x1, 1);
		g[y1].emplace_back(y0, 0);
		g[y0].emplace_back(y1, 1);
	}
 
	deque<PI> q;
	q.emplace_front(P(1, 0), 0);
	map<P, int> minc;
	while (!q.empty()) {
		PI pi = q.front();
		q.pop_front();
		if (minc.find(pi.first) != minc.end() && minc[pi.first] <= pi.second) continue;
		minc[pi.first] = pi.second;
		if (pi.first == P(n, 0)) break;
		for (PI& npi : g[pi.first]) {
			int cost = pi.second + npi.second;
			if (minc.find(npi.first) != minc.end() && minc[npi.first] <= cost) continue;
			if (npi.second) q.emplace_back(npi.first, cost);
			else q.emplace_front(npi.first, cost);
		}
	}
	P dst(n, 0);
	if (minc.find(dst) == minc.end()) cout << -1 << endl;
	else cout << minc[dst] << endl;
	return 0;
}