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;

#define int long long

int dx[] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};

int N, M;
int p[210000], q[210000], c[210000];
multimap<int, int> G[110000];
int d[110000];

signed main() {
    cin >> N >> M;
    for (int i = 0; i < M; i++) {
        cin >> p[i] >> q[i] >> c[i];
        G[p[i]].emplace(c[i], q[i]);
        G[q[i]].emplace(c[i], p[i]);
    }
    fill(d, d + N + 1, -1);
    d[1] = 0;
    queue<int> bfs;
    bfs.push(1);
    while (!bfs.empty()) {
        int cur = bfs.front();
        bfs.pop();
        while (!G[cur].empty()) {
            int color = G[cur].begin()->first;
            stack<int> dfs;
            dfs.push(cur);
            while (!dfs.empty()) {
                int curcur = dfs.top();
                dfs.pop();
                auto p = G[curcur].equal_range(color);
                for (auto it = p.first; it != p.second; ++it) {
                    int next = it->second;
                    if (d[next] == -1) {
                        bfs.push(next);
                        d[next] = d[cur] + 1;
                    }
                    dfs.push(next);
                }
                G[curcur].erase(color);
            }
        }
    }
    cout << d[N] << endl;
}