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;
 
//#pragma GCC optimize("O3")
//#pragma GCC target("sse4")
 
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define RFOR(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define FILL(a, value) memset(a, value, sizeof(a))
 
#define SZ(a) (int)((a).size())
#define ALL(a) a.begin(), a.end()
 
typedef long long LL;
typedef pair<int, int> PII;

const int MAX = 1 << 17;

vector<PII> g[MAX];
int used[MAX];

int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	int n, m;
	cin >> n >> m;
	FOR(i, 0, m)
	{
		int u, v, w;
		cin >> u >> v >> w;
		u--;
		v--;
		g[u].push_back({v, w});
		g[v].push_back({u, w});
	}
	map<PII, int> d;
	deque<PII> Q;
	d[{0, 0}] = 0;
	Q.push_back({0, 0});
	set<PII> upd;
	int ans = n;
	while(!Q.empty())
	{
		auto vv = Q.front();
		Q.pop_front();
		int v = vv.first;
		int dvv = d[vv];
		if(v == n - 1)
			ans = min(ans, dvv);
		if(upd.count(vv))
			continue;
		upd.insert(vv);
		used[v]++;
		if(used[v] <= 2)
		{
			for(auto to: g[v])
			{
				int dto = dvv + (vv.second != to.second);
				if(!d.count(to) || dto < d[to])
				{
					d[to] = dto;
					if(vv.second == to.second)
						Q.push_front(to);
					else
						Q.push_back(to);
				}
			}
		}
	}
	cout << (ans == n ? -1 : ans) << "\n";
	return 0;
}