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 For( i, j, k ) for( int i = j; i <= k; ++i )
using std::cin;
using std::cout;
using std::vector;
using std::pair;
using std::string;
const int Maxn = 1e5 + 1000;
int n, m, p, q, c, dis[ Maxn ];
vector < pair < int, int > > v[ Maxn ];
std::set < int > elvahs[ Maxn ];
bool vis[ Maxn ];
using std::make_pair;
int main( )
{
	std::ios::sync_with_stdio( false );
	cin >> n >> m;
	For( i, 1, m )
	{
		cin >> p >> q >> c;
		v[ q ].push_back( make_pair( p, c ) );
		v[ p ].push_back( make_pair( q, c ) );
	}
	std::priority_queue < pair < int, int >, vector < pair < int, int > >, std::less < pair < int, int > > > Q;
	std::memset( dis, 63, sizeof dis );
	Q.push( make_pair( dis[ 1 ] = 0, 1 ) );
	while( !Q.empty( ) )
	{
		auto tp = Q.top( );
		Q.pop( );
		if( vis[ tp.second ] )
			continue;
		vis[ tp.second ] = 1;
		for( auto &i: v[ tp.second ] )
		{
			bool k = elvahs[ tp.second ].count( i.second );
			if( !k + dis[ tp.second ] < dis[ i.first ] )
			{
				dis[ i.first ] = !k + dis[ tp.second ];
				elvahs[ i.first ].clear( );
				Q.push( make_pair( -dis[ i.first ], i.first ) );
				elvahs[ i.first ].insert( i.second );
			}
			else if( !k + dis[ tp.second ] == dis[ i.first ] )
				elvahs[ i.first ].insert( i.second );
		}
	}
	cout << ( dis[ n ] == 0x3f3f3f3f ? -1 : dis[ n ] );
	return 0;
}