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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
//T+N
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
#define endl "\n"
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template<class T, class T2> inline bool chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline bool chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const ll mod = 1e9 + 7;
template<class T> inline void fix(T &x) {if(x >= mod | x <= -mod) {x %= mod;} if(x < 0) {x += mod;}}
#define out(x) cout << __LINE__ << ": " << (#x) << " = " << (x) << endl

const int MAX_N = 4e5 + 10;
vector<int > g[MAX_N];

ll dist[MAX_N], dest[MAX_N], tpe[MAX_N];
ll ans, n;

void bfs() {
    fill_n(dist, MAX_N, mod);
    deque<pair<int, int> > q;
    q.push_back({0, 0});
    dist[0] = 0;
    while(!q.empty()) {
        auto now = q.front(); q.pop_front();
        int curr = now.first;
        if(now.second != dist[curr]) {continue;}
        if(dest[curr] == n) {
            chkmin(ans, dist[curr]);
        }
        for(auto it : g[dest[curr]]) {
            if(tpe[it] == tpe[curr]) {
                if(dist[curr] < dist[it]) {
                    dist[it] = dist[curr];
                    q.push_front({it, dist[it]});
                }
            } else {
                if(dist[curr] + 1 < dist[it]) {
                    dist[it] = dist[curr] + 1;
                    q.push_back({it, dist[it]});
                }
            }
        }
    }
}

signed main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
    int m;
    cin >> n >> m;
    tpe[0] = -1;
    dest[0] = 1;
    for(int i = 1; i <= m; i ++) {
        int a, b, c;
        cin >> a >> b >> c;
        g[a].push_back(2 * i);
        g[b].push_back(2 * i + 1);
        tpe[2 * i] = c;
        tpe[2 * i + 1] = c;
        dest[2 * i] = b;
        dest[2 * i + 1] = a;
    }
    ans = mod;
    bfs();
    if(ans == mod) {
        cout << -1 << endl;
    } else {
        cout << ans << endl;
    }
    return 0;
}