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)

#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,no-stack-protector,fast-math")
#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;
const int N = 2e5 + 5, M = 2 * N + 5;

int head[N], nxt[M], to[M], typ[N], ne, n;

void init(){
    ne = 0;
    memset(head, -1, n * sizeof head[0]);
}

void addEdge(int f, int t, int ty){
    to[ne] = t;
    typ[ne] = ty;
    nxt[ne] = head[f];
    head[f] = ne++;
}

void addBiEdge(int a, int b, int ty){
    addEdge(a, b, ty);
    addEdge(b, a, ty);
}

int dis[N];

int bfs(){
    queue<pair<int, int> > q;
    q.emplace(0, -1);

    memset(dis, '?', n * sizeof dis[0]);
    dis[0] = 0;

    while(q.size()){
        int u = q.front().first;
        if(u == n - 1)return dis[u];

        int lst = q.front().second;
        q.pop();

        for(int e = head[u] ; ~e ; e = nxt[e]){
            int v = to[e];
            int cst = (typ[e] != lst);
            if(dis[u] + cst < dis[v]){
                dis[v] = dis[u] + cst;
                q.emplace(v, typ[e]);
            }
        }

    }

    return -1;
}

int main(){
    int m;
    scanf("%d %d", &n, &m);
    init();
    while(m--){
        int u, v, ty;
        scanf("%d %d %d", &u, &v, &ty);
        addBiEdge(--u, --v, ty);
    }
    printf("%d\n", bfs());
}