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>
typedef long long LL;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;

int n, m;
LL zip(LL p, LL c) {
  return n * c + p;
}

int main(void)
{
  cin >> n >> m;
  map<LL, set<tuple<LL, int> >> e;
  REP(i,m) {
    int p, q, c;
    cin >> p >> q >> c;
    e[zip(p, c)].emplace(zip(q, c), 0);
    e[zip(q, c)].emplace(zip(p, c), 0);
    e[zip(p, 0)].emplace(zip(p, c), 1);
    e[zip(p, c)].emplace(zip(p, 0), 0);
    e[zip(q, 0)].emplace(zip(q, c), 1);
    e[zip(q, c)].emplace(zip(q, 0), 0);
  }
  set<LL> already;
  priority_queue<tuple<int, LL>> pq; // negc, station
  pq.emplace(0, zip(1, 0));
  while(!pq.empty()) {
    int negc;
    LL station;
    tie(negc, station) = pq.top();
    pq.pop();
    if(!already.insert(station).second) continue;
    if(station == zip(n,0)) {
      cout << -negc << endl;
      return 0;
    }
    for(auto t : e[station]) {
      LL to;
      int ec;
      tie(to, ec) = t;
      pq.emplace(negc - ec, to);
    }
  }
  cout << -1 << endl;
  return 0;
}