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;
int n, m, x, y, c, cnt;
int d[500020];
vector<pair<int, int> > a[500020];
map<int, int> g[100020];
deque<int> q;
int get(int x, int y) {
	if (g[x].count(y)) {
		return g[x][y];
	}
	return g[x][y] = ++cnt;
}
void add(int x, int y, int z) {
	a[x].push_back(make_pair(y, z));
	a[y].push_back(make_pair(x, z));
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) {
		get(i, 0);
	}
	for (int i = 0; i < m; i++) {
		scanf("%d%d%d", &x, &y, &c);
		int x0 = get(x, 0), xc = get(x, c);
		int y0 = get(y, 0), yc = get(y, c);
		add(xc, yc, 0);
		add(x0, xc, 1);
		add(yc, y0, 1);
	}
	memset(d, -1, sizeof d);
	d[1] = 0;
	q.push_back(1);
	while (q.size()) {
		int u = q.back();
		q.pop_back();
		for (pair<int, int> i: a[u]) {
			if (d[i.first] == -1 || d[i.first] > d[u] + i.second) {
				d[i.first] = d[u] + i.second;
				if (i.second == 1) {
					q.push_front(i.first);
				} else {
					q.push_back(i.first);
				}
			}
		}
	}
	printf("%d\n", d[n] >> 1);
	return 0; 
}