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 IL inline 
IL int read() {
	char ch = getchar(); int u = 0, f = 1;
	while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
	while (isdigit(ch)) { u = (u << 1) + (u << 3) + ch - 48; ch = getchar(); }
	return u * f; 
}

const int maxn = 5e5 + 10;
int dis[maxn]; bool flag[maxn];
set<int>F[maxn];
vector<pair<int,int> >e[maxn];

int main() {
#ifdef Edgration
	freopen("1.in", "r", stdin);
#endif
	int n = read(), m = read();
	for (int i = 1; i <= m; ++i) {
		int x = read(), y = read(), c = read();
		e[x].push_back(make_pair(y, c)); e[y].push_back(make_pair(x, c));
	}	
	priority_queue<pair<int,int> >q;
	q.push(make_pair(0, 1)); 
	memset(dis, 0x3f, sizeof dis); dis[1] = 0;
	while (!q.empty()) {
		int u = q.top().second; q.pop();
		if (flag[u]) continue;
		flag[u] = 1;
		for (int i = 0; i < e[u].size(); ++i) {
			int v = e[u][i].first; int w = e[u][i].second;
			int c = F[u].count(w) ? 0 : 1;
			if (dis[v] > dis[u] + c) {
				dis[v] = dis[u] + c;
				F[v].clear();
				F[v].insert(w);
				q.push(make_pair(-dis[v], v));
			}
			else if (dis[v] == dis[u] + c) F[v].insert(w);
		}
	}
	cout << (dis[n] == 0x3f3f3f3f ? -1 : dis[n]) << endl;
	return 0;
}