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 for_(i,a,b) for(int i=(a);i<(b);++i)
typedef long long lint;
template< typename T > using Vec = vector< T >;

lint encode(lint v, lint c) { return (v << 32) | c; }

struct Edge {
	lint to;
	int dst;
};

int N, M;
unordered_map< lint, Vec< Edge > > edges;
unordered_map< lint, int > dst;

void solve() {
	deque< lint > que;
	que.push_back(0);
	dst[0] = 0;
	
	while (!que.empty()) {
		lint cur = que.front(); que.pop_front();
				
		for (const Edge& e : edges[cur]) {
			if (dst.count(e.to) == 0 || dst[e.to] > dst[cur] + e.dst) {
				dst[e.to] = dst[cur] + e.dst;
				if (e.dst == 0) que.push_front(e.to);
				else que.push_back(e.to);
			}
		}
	}
	
	if (dst.count(encode(N-1, 0)) == 0) puts("-1");
	else cout << dst[encode(N-1, 0)] << endl;
}

int main() {
	cin >> N >> M;
	
	for_(i,0,M) {
		int u, v, c;
		cin >> u >> v >> c;
		--u; --v;
		
		edges[encode(u, c)].push_back(Edge{encode(v, c), 0});
		edges[encode(v, c)].push_back(Edge{encode(u, c), 0});
		
		edges[encode(u, 0)].push_back(Edge{encode(u, c), 1});
		edges[encode(v, 0)].push_back(Edge{encode(v, c), 1});
		
		edges[encode(u, c)].push_back(Edge{encode(u, 0), 0});
		edges[encode(v, c)].push_back(Edge{encode(v, 0), 0});
	}
	
	solve();
}