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>
#define sc(x) scanf("%d", &x)
#define pr(x) printf("%d\n", x)
#define lowbit(x) (x & -x)
#define ll long long
#define N 500010
using namespace std;
ll tot = 0, head[N], v[N], dis[N], n, m, cnt;
struct Xiao
{
	int next, to, w;
}f[N * 3];
void add(int x, int y, int z)
{
	f[++tot].to = y;
	f[tot].next = head[x];
	f[tot].w = z;
	head[x] = tot;
}
void spfa()
{
	memset(dis, 127, sizeof dis);
	memset(v, 0, sizeof v);
	queue <int> q;
	q.push(1);
	v[1] = 1;
	dis[1] = 0;
	while (q.size())
	{
		int x = q.front();
		q.pop();
		for (int i = head[x]; i; i = f[i].next)
		{
			int y = f[i].to;
			if (dis[y] > dis[x] + f[i].w)
			{
				dis[y] = dis[x] + f[i].w;
				if (!v[y])
				{
					v[y] = 1;
					q.push(y);
				}
			}
		}
		v[x] = 0;
	}
}
map < pair <int, int>, int > p;
int get(int x, int y)
{
	if (p.find(make_pair(x, y)) != p.end()) return p[make_pair(x, y)];
	else return p[make_pair(x, y)] = ++cnt;
}
int main()
{
	//freopen("railway.in", "r", stdin);
	//freopen("railway.out", "w", stdout);
	sc(n);
	sc(m);
	cnt = n;
	for (int i = 1; i <= m; i++)
	{
		int x, y, z;
		sc(x);
		sc(y);
		sc(z);
		int p1 = get(x, z);
		int p2 = get(y, z);
		add(p2, p1, 0);
		add(p1, p2, 0);
		add(x, p1, 1);
		add(p1, x, 1);
		add(y, p2, 1);
		add(p2, y, 1);
	}
	spfa();
	printf("%lld\n", dis[n] > 2 * n ? -1 : dis[n] / 2);
	return 0;
}