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 int64 = long long;
const int INF = 1 << 30;


int N, M;
map< int, vector< int > > g[1000000];
vector< int > add[100000];
int v[100000];

int main() {
  cin >> N >> M;
  for(int i = 0; i < M; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    --a, --b, --c;
    g[c][a].push_back(b);
    g[c][b].push_back(a);
    add[a].push_back(c);
    add[b].push_back(c);
  }
  queue< int > que;
  que.emplace(0);
  memset(v, -1, sizeof(v));
  v[0] = 0;
  set< pair< int, int > > eval;
  function< void(int, int, int) > dfs = [&](int c_idx, int idx, int cur) {
    if(eval.count({c_idx, idx})) return;
    eval.emplace(c_idx, idx);
    for(auto &to : g[c_idx][idx]) {
      dfs(c_idx, to, cur);
      if(v[to] == -1) {
        v[to] = cur;
        que.emplace(to);
      }
    }
  };
  while(que.size()) {
    int p = que.front();
    que.pop();
    for(auto &luz : add[p]) dfs(luz, p, v[p] + 1);
  }
  cout << v[N - 1] << endl;
}