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 all(x) x.begin(), x.end()
using namespace std;
using ll = long long;
using ld = long double;
const int maxn = 1<<18;
int n, m, d[maxn], vis[maxn], t = 1;
vector<array<int, 2>> g[maxn];
map<int, vector<int>> gc[maxn];
queue<int> pq;
set<int> cv[maxn];
void enq(int v, int c, int x) {
	if(!cv[v].insert(c).second) return;
	if(d[v] > x) d[v] = x, pq.push(v);
	for(auto i : gc[v][c]) {
		enq(i, c, x);
	}
}
int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m;
	for(int f, t, c, i = 0; i < m; i++) {
		cin >> f >> t >> c;
		g[f].push_back({t, c});
		g[t].push_back({f, c});
		gc[f][c].push_back(t);
		gc[t][c].push_back(f);
	}
	memset(d, 0x3f, sizeof d);
	d[1] = 0;
	int v;
	pq.push(1);
	while(!pq.empty()) {
		t++;
		v = pq.front(); pq.pop();
		for(auto _i : g[v]) {
			int i = _i[0], c = _i[1];
			enq(i, c, d[v]+1);
		}
	}
	if(d[n] > n) d[n] = -1;
	cout << d[n];
}