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 <unordered_set>
#include <unordered_map>
#include <queue>
#include <functional>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
typedef long long ll;
using namespace std;
int main() {
    // input
    int n, m; cin >> n >> m;
    vector<unordered_map<int,vector<int> > > g(n);
    repeat (i,m) {
        int p, q, c; cin >> p >> q >> c; -- p; -- q;
        g[p][c].push_back(q);
        g[q][c].push_back(p);
    }
    // compute
    vector<int> dist(n, -1);
    queue<int> que;
    dist[0] = 0;
    que.push(0);
    vector<unordered_set<int> > used(n);
    function<void (int, int, int)> dfs = [&](int i, int c, int d) {
        used[i].insert(c);
        if (dist[i] == -1) {
            dist[i] = d;
            que.push(i);
        }
        for (int j : g[i][c]) {
            if (used[j].count(c)) continue;
            if (dist[j] != -1 and dist[j] < d) continue;
            dfs(j, c, d);
        }
    };
    while (not que.empty()) {
        int i = que.front(); que.pop();
        for (auto it : g[i]) {
            int c = it.first;
            if (not used[i].count(c)) {
                dfs(i, c, dist[i]+1);
            }
        }
    }
    // output
    cout << dist[n-1] << endl;
    return 0;
}