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 MOD @
#define ADD(X,Y) ((X) = ((X) + (Y)%MOD) % MOD)
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;
 
int N, M;
map<pair<int, int>, int> stnid;
int idlast;
vector<pair<int, int> > graph[606060];
bool vis[606060];
 
void ae(int p, int q, int c, int d)
{
	graph[p].push_back({ q, c });
	graph[q].push_back({ p, d });
}
 
int getid(int stn, int cmp)
{
	if (stnid.count({ stn, cmp })) return stnid[{stn, cmp}];
	int ret = idlast++;
	ae(stn, ret, 1, 0);
	return stnid[{stn, cmp}] = ret;
}
 
int main()
{
	scanf("%d%d", &N, &M);
	idlast = N;
	for (int i = 0; i < M; ++i) {
		int p, q, c;
		scanf("%d%d%d", &p, &q, &c);
		--p; --q;
 
		int a = getid(p, c);
		int b = getid(q, c);
		ae(a, b, 0, 0);
	}
 
	priority_queue<pair<int, int> > Q;
	for (int i = 0; i < idlast; ++i) vis[i] = false;
	Q.push({ 0, 0 });
	while (!Q.empty()) {
		pair<int, int> tmp = Q.top(); Q.pop();
		int cost = -tmp.first;
		int pt = tmp.second;
		if (vis[pt]) continue;
		vis[pt] = true;
		if (pt == N - 1) {
			printf("%d\n", cost);
			return 0;
		}
		for (auto e : graph[pt]) {
			Q.push({ -(cost + e.second), e.first });
		}
	}
	puts("-1");
	return 0;
}