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: Accepted)

#include <bits/stdc++.h>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn = 2e6 + 10;
using namespace std;
int n, m, k, t, h[maxn], tot, ans[maxn];
set<int>comp[maxn];
struct node
{
        int to, nxt, com;
} e[maxn << 2];
void add(int x, int y, int z)
{
        tot++;
        e[tot].to = y;
        e[tot].com = z;
        e[tot].nxt = h[x];
        h[x] = tot;
}
priority_queue<pair<int, int> >p;
int main()
{
        int i, j;
        while (scanf("%d %d", &n, &m) == 2)
        {
                //cout << n << " " << m << endl;
                for (int i = 1; i <= n; i++)
                {
                        h[i] = 0;
                        ans[i] = inf;
                        comp[i].clear();
                }
                tot = 0;
                while (m--)
                {
                        int a, b, c;
                        scanf("%d%d%d", &a, &b, &c);
                        add(a, b, c);
                        add(b, a, c);
                }
                p.push({0, 1});
                ans[1] = 0;
                while (!p.empty())
                {
                        int now = p.top().se, ca = -p.top().fi;
                        p.pop();
                        if (ans[now] < ca)
                        {
                                continue;
                        }
                        for (i = h[now]; i; i = e[i].nxt)
                        {
                                int to = e[i].to, to_com = e[i].com;
                                int to_ca = ca + (!comp[now].count(to_com));
                                if (ans[to] > to_ca)
                                {
                                        ans[to] = to_ca;
                                        p.push({ -to_ca, to});
                                        comp[to].clear();
                                        comp[to].insert(to_com);
                                }
                                else if (ans[to] == to_ca)
                                {
                                        comp[to].insert(to_com);
                                }
                        }
                }
                printf("%d\n", ans[n] == inf ? -1 : ans[n]);
        }
        //system("Pause");
        return 0;
}