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;
using edges = vector<int>;
using graph = vector<unordered_map<int, edges>>;
 
const int INF = 1e9;
//
graph G;
vector<int> d;
vector<unordered_set<int>> used;
queue<int> q;
void dfs(int i, int c, int dist) {
  used[i].insert(c);
  if (d[i] == -1) {
    d[i] = dist;
    q.push(i);
  }
  for (auto&& j : G[i][c]) {
    if (used[j].count(c)) continue;
    if (d[j] != -1 && d[j] < dist) continue;
    dfs(j, c, dist);
  }
}
void dijkstra(int s) {
  int N = G.size();
  d.assign(N, -1); d[s] = 0;
  used.resize(N);
  q.push(s);
  while (!q.empty()) {
    int i = q.front(); q.pop();
    for (auto&& itr : G[i]) {
      int c = itr.first;
      if (!used[i].count(c)) dfs(i, c, d[i] + 1);
    }
  }
}
 
int main() {
  int N, M; cin >> N >> M;
  G.resize(N);
  for (int i = 0; i < M; ++i) {
    int p, q, c; cin >> p >> q >> c;
    --p; --q;
    G[p][c].push_back(q);
    G[q][c].push_back(p);
  }
  dijkstra(0);
  cout << d[N - 1] << endl;
  return 0;
}