Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std ;

#define N 200010
#define inf 0x3f3f3f3f

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

int n , m ; 
int head[ N << 2 ] , cnt ;
int d[ N ] , now[ N ] , vis[ N ] ;
int q[ N * 100 ] ;

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

void spfa() {
	for( int i = 1 ; i <= n ; i ++ ) d[ i ] = inf ;
	q[ 1 ] = vis[ 1 ] = 1 ;
	d[ 1 ] = 0 ;
	int l = 1 , r = 2 ;
	while( l < r ) {
		int u = q[ l ++ ] ;
		vis[ u ] = 0 ;
		for( int i = head[ u ] ; i ; i = e[ i ].nxt ) {
			int v = e[ i ].to ;
			if( now[ u ] == e[ i ].belong && d[ v ] > d[ u ] ) {
				d[ v ] = d[ u ] ;
				now[ v ] = e[ i ].belong ;
				if( ! vis[ v ] ) vis[ v ] = 1 , q[ r ++ ] = v ;
			}
			if(  now[ u ] != e[ i ].belong && d[ v ] > d[ u ] + 1 ) {
				d[ v ] = d[ u ] + 1 ;
				now[ v ] = e[ i ].belong ;
				if( ! vis[ v ] ) vis[ v ] = 1 , q[ r ++ ] = v ;
			}
		}
	}
	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() ;
}