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;

const int N = 1e5 + 10;

int n, m, cnt, d[N << 3], v[N << 3], sum, ans;

map <int, int> mp[N << 3];

vector < pair <int, int> > e[N << 3];

queue <int> q;

void init ()
{
	scanf ( "%d%d" , &n, &m);
	cnt = n;
	for (int i = 1; i <= m; ++ i)
	{
		int p, q, c;
		scanf ( "%d%d%d" ,&p, &q, &c);
		if ( !mp[p].count (c) )
		{
			mp[p][c] = ++cnt;
			e[p].push_back ( make_pair (cnt, 1) );
			e[cnt].push_back ( make_pair (p, 1) );
		}
		if ( !mp[q].count (c) )
		{
			mp[q][c] = ++cnt;
			e[q].push_back ( make_pair (cnt, 1) );
			e[cnt].push_back ( make_pair (q, 1) );
		}
		e[ mp[p][c] ].push_back ( make_pair (mp[q][c], 0) );
		e[ mp[q][c] ].push_back ( make_pair (mp[p][c], 0) );
	}
}

void work ()
{
	memset ( d, 0x3f, sizeof (d) );
	d[1] = 0, v[1] = 1;
	q.push (1);
	for ( ; !q.empty (); )
	{
		int x = q.front ();
		q.pop ();
		for (int i = 0; i < e[x].size (); ++ i)
		{
			int y = e[x][i].first;
			if (d[y] > d[x] + e[x][i].second)
			{
				d[y] = d[x] + e[x][i].second;
				if (!v[y])
				{
					v[y] = 1;
					q.push (y);
				}
			}
		}
		v[x] = 0;
	}
	if (d[n] == 0x3f3f3f3f) sum = -1;
	else sum = d[n] / 2;
}

void prin ()
{
	printf ( "%d" , sum);
}

int main ()
{
	init ();
	work ();
	prin ();
	return 0;
}