Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <cstdio>
#include <utility>
#include <vector>
static const int MAXN = 1e5 + 3;
static const int MAXM = 2e5 + 3;
static const int MAXC = 1e6 + 2;
static const int INF = 0x3fffffff;

int n, m;
std::vector<std::pair<int, int>> c[MAXC];

int ey_sz;
std::vector<int> ex[MAXN], ey[MAXN + MAXM];

int bfs_ct = 0;
int vis[MAXN] = { 0 };
inline void bfs_renew() { bfs_ct++; }
inline void bfs_x(int u)
{
    if (vis[u] == bfs_ct) return;

    static int q[MAXN], qhead, qtail;
    qhead = 0, qtail = 1;
    vis[u] = bfs_ct; q[0] = u;

    while (qhead < qtail) {
        u = q[qhead++];
        for (int v : ex[u]) if (vis[v] != bfs_ct) {
            vis[v] = bfs_ct; q[qtail++] = v;
        }
    }

    //for (int i = 0; i < qtail; ++i) printf(" %d", q[i]); putchar('\n');
    int idx = ey_sz++;
    for (int i = 0; i < qtail; ++i) {
        ey[q[i]].push_back(idx);
        ey[idx].push_back(q[i]);
    }
}

int dist[MAXN];
inline void bfs_y(int u)
{
    for (int i = 0; i < ey_sz; ++i) dist[i] = INF;
    static int q[MAXN], qhead, qtail;
    qhead = 0, qtail = 1;
    vis[u] = bfs_ct; q[0] = u; dist[u] = 0;

    while (qhead < qtail) {
        u = q[qhead++];
        for (int v : ey[u]) if (vis[v] != bfs_ct) {
            vis[v] = bfs_ct; q[qtail++] = v; dist[v] = dist[u] + 1;
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    int u, v, w;
    for (int i = 0; i < m; ++i) {
        scanf("%d%d%d", &u, &v, &w); --u, --v, --w;
        c[w].push_back(std::make_pair(u, v));
    }

    std::vector<int> verts;
    ey_sz = n;
    for (int i = 0; i < MAXC; ++i) if (!c[i].empty()) {
        for (auto e : c[i]) {
            ex[e.first].push_back(e.second);
            ex[e.second].push_back(e.first);
            verts.push_back(e.first);
            verts.push_back(e.second);
        }
        bfs_renew();
        for (int u : verts) bfs_x(u);
        for (int u : verts) ex[u].clear();
        verts.clear();
    }

    bfs_renew();
    bfs_y(0);
    printf("%d\n", dist[n - 1] == INF ? -1 : dist[n - 1] / 2);

    return 0;
}