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 <iostream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <deque>
#include <algorithm>
#include <numeric>
using namespace std;
using ll = long long;
using V = pair<int, int>;
using E = pair<V, int>;
 
int main() {
  int N, M;
  cin >> N >> M;
  map<V, vector<E>> G;
  for (int i = 0; i < M; i++) {
    int p, q, c;
    cin >> p >> q >> c;
    V v1 = V(p, c);
    V v2 = V(q, c);
    G[v1].push_back(E(v2, 0));
    G[v2].push_back(E(v1, 0));
  }
  map<V, vector<E>> newG = G;
  for (auto &ps: G) {
    V v1 = ps.first;
    V v2 = V(v1.first, -1);
    newG[v1].push_back(E(v2, 0));
    newG[v2].push_back(E(v1, 1));
  }
  G = newG;
  map<V, int> dist;
  priority_queue<pair<int, V>> q;
  V start = V(1, -1), goal = V(N, -1);
  dist[start] = 0;
  q.emplace(0, start);
  while (!q.empty()) {
    auto ps = q.top(); q.pop();
    int cost = -ps.first;
    V v = ps.second;
    if (dist[v] < cost) {
      continue;
    }
    for (auto e: G[v]) {
      V u = e.first;
      int add = e.second;
      if (!dist.count(u) || cost + add < dist[u]) {
        dist[u] = cost + add; 
        q.emplace(-dist[u], u);
      }
    }
  }
  cout << (dist.count(goal) ? dist[goal] : -1) << endl;
  return 0;
}