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 ;

#define N 500010
#define inf 0x3f3f3f3f

struct node {
	int to , nxt , belong ;
}e[ N << 2 ];

int n , m ; 
int head[ N << 2 ] , cnt ;


int d[ N * 2 ]  ;

void ins( int u , int v , int c ) {
	e[ ++ cnt ].to = v ;
	e[  cnt ].nxt = head[ u ] ;
	e[ cnt ].belong = c ;
	head[ u ] = cnt ;
}

std::set< int > st[ N ] ;
priority_queue< pair<int,int> > q;

void spfa() {
	for( int i = 1 ; i <= n ; i ++ ) d[ i ] = inf ;
	d[ 1 ] = 0 ;
	q.push( { 0 , 1 } ) ;
	while(!q.empty()){
		int u = q.top().second , ans = -q.top().first;
		q.pop() ;
		if( d[ u ] < ans ) continue ;
		for( int i = head[ u ] ; i ; i =e[ i ].nxt ) {
			int v = e[ i ].to ;
			int num = ans + !st[ u ].count( e[ i ].belong ) ;
			if( d[ v ] > num ) {
				d[ v ] = num ;
				q.push( { -num , v } ) ;
				st[ v ].clear() ;
				st[ v ].insert( e[ i ].belong ) ;
			}else if( d[ v ] == num ) {
				st[ v ].insert( e[ i ].belong ) ;
			}
		}
	}
	if( d[ n ] == inf ) puts( "-1" ) ;
	else printf( "%d\n" , d[ n ] ) ;
}

int main() {
	scanf( "%d%d" , &n ,&m ) ;
	for( int i = 1 ; i <= m ; i ++ ) {
		int x , y , c ;
		scanf( "%d%d%d" , &x , &y , &c ) ;
		ins( x , y , c ) ;
		ins( y , x , c ) ;
	}
	spfa() ;
}