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, d[500005], cnt;
map<int, int> g[500005];
deque<int> q;
vector<pair<int, int> > E[500005];
int get(int x, int y) {
	if (g[x].count(y)) {
		return g[x][y];
	}
	g[x][y] = ++cnt;
	E[cnt].clear();
	return cnt;
}
void add(int u, int v, int w) {
	E[u].push_back(make_pair(v, w));
	E[v].push_back(make_pair(u, w));
}
int main() {
	while (~scanf("%d%d", &n, &m)) {
		cnt = 0;
		for (int i = 1; i <= n; ++i)
			g[i].clear();
		for (int i = 1; i <= n; ++i)
			get(i, 0);
		for (int i = 1; i <= m; ++i) {
			int x, y, c;
			scanf("%d%d%d", &x, &y, &c);
			int x0 = get(x, 0), y0 = get(y, 0);
			int xc = get(x, c), yc = get(y, c);
			add(x0, xc, 1);
			add(y0, yc, 1);
			add(xc, yc, 0);
		}
		memset(d, -1, sizeof(d));
		d[1] = 0;
		q.push_back(1);
		while (!q.empty()) {
			int u = q.front();
			q.pop_front();
			for (int i = 0; i < E[u].size(); ++i) {
				pair<int, int> e = E[u][i];
				if (d[e.first] == -1 || d[e.first] > d[u] + e.second) {
					d[e.first] = d[u] + e.second;
					if (e.second) {
						q.push_back(e.first);
					} else
						q.push_front(e.first);
				}
			}
		}
		printf("%d\n", d[n] >> 1);
	}
	return 0;
}