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;

typedef long long ll;
const int N = 100005;
const int M = 1000000007;

int n, m, tot;
map<pair<int, int>, int> id;

typedef int wtype;
struct Node {
  int v; wtype w;
  bool operator<(const Node &r)const { return w > r.w; }
};
vector<Node> e[N * 22]; wtype dist[N * 22];
void Dijkstra(int src) {
  memset(dist, 0x3f, sizeof(dist));
  priority_queue<Node> que; que.push((Node) {src, dist[src] = 0});
  while (!que.empty()) {
    int u = que.top().v; wtype w = que.top().w; que.pop();
    if (w > dist[u]) { continue; }
    for (int i = 0; i < (int)e[u].size(); i++) {
      int v = e[u][i].v; wtype w = e[u][i].w;
      if (dist[v] > dist[u] + w) {
        dist[v] = dist[u] + w; que.push((Node) {v, dist[v]});
      }
    }
  }
}

inline int get(int x, int y) {
  if (id.find({x, y}) != id.end()) { return id[ {x, y}]; }
  e[x].push_back({tot, 1}); e[tot].push_back({x, 0});
  return id[ {x, y}] = tot++;
}

int main() {
  scanf("%d%d", &n, &m);
  tot = n + 1;
  for (int i = 0, u, v, c; i < m; i++) {
    scanf("%d%d%d", &u, &v, &c);
    int uid = get(u, c), vid = get(v, c);
    e[uid].push_back({vid, 0}); e[vid].push_back({uid, 0});
  }
  Dijkstra(1);
  printf("%d\n", dist[n] == 0x3f3f3f3f ? -1 : dist[n]);
}