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: Wrong Answer)

#include <bits/stdc++.h>
#define For(i, a, b) for(int (i)=(a); (i)<(b); ++(i))
#define rFor(i, a, b) for(int (i)=(a)-1; (i)>=(b); --(i))
#define rep(i, n) For((i), 0, (n))
#define rrep(i, n) rFor((i), (n), 0)
#define fi first
#define se second
using namespace std;
typedef long long lint;
typedef pair<int, int> pii;

typedef struct UnionFindTree{
    vector<int> par;
    
    UnionFindTree(int n){
        par.resize(n, -1);
    }
    
    int find(int x){
        if(par[x] < 0) return x;
        return par[x] = find(par[x]);
    }
    
    int size(int x){
        return -par[find(x)];
    }
    
    bool unite(int x, int y){
        x = find(x);
        y = find(y);
        if(x == y) return false;
        if(size(x) < size(y)) swap(x, y);
        par[x] += par[y];
        par[y] = x;
        return true;
    }
    
    bool same(int x, int y){
        return find(x) == find(y);
    }
}UF;
 
template<class T> struct edge{
    int from, to; T cost;
    edge(int f, int t, T c){
        from = f; to = t; cost = c;
    }
};
 
template<class T> struct Graph{
    vector<vector<edge<T>>> G;
    int n;
    
    Graph(int n_){
        n = n_;
        G.resize(n);
    }
    
    void add_edge(int f, int t, T c){
        G[f].emplace_back(f, t, c);
    }
    
    pair<bool, vector<T>> bellman_ford(int s){
        T INF = numeric_limits<T>::max();
        vector<T> d(n, INF);
        vector<edge<T>> E;
        rep(i, n)for(edge<T> &e: G[i]) E.push_back(e);
        d[s] = 0;
        rep(i, n)for(edge<T> &e: E){
            if(d[e.from] != INF && d[e.from] + e.cost < d[e.to]){
                d[e.to] = d[e.from] + e.cost;
                if(i == n-1) return make_pair(true, d);
            }
        }
        return make_pair(false, d);
    }
    
    vector<T> dijkstra(int s){
        using P = pair<T, int>;
        priority_queue<P, vector<P>, greater<P>> que;
        vector<T> d(n, numeric_limits<T>::max());
        d[s] = 0;
        que.push(P((T)0, s));
        while(!que.empty()){
            P p = que.top(); que.pop();
            int v = p.second;
            if(d[v] < p.first) continue;
            for(edge<T> &e : G[v]){
                if(d[e.to] > d[v] + e.cost){
                    d[e.to] = d[v] + e.cost;
                    que.push(P(d[e.to], e.to));
                }
            }
        }
        return d;
    }
    
    pair<bool, vector<vector<T>>> warshall_floyd(){
        T INF = numeric_limits<T>::max();
        vector<vector<T>> d = vector<vector<T>>(n, vector<T>(n, INF));
        rep(i, n){
            for(edge<T> &e: G[i]) d[i][e.to] = e.cost;
            d[i][i] = 0;
        }
        rep(k, n)rep(i, n)rep(j, n)if(d[i][k] < INF && d[k][j] < INF){
            d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
        }
        rep(i, n)if(d[i][i] < 0) return make_pair(true, d);
        return make_pair(false, d);
    }
    
    T kruskal(){
        vector<edge<T>> E;
        rep(i, n)for(edge<T> &e: G[i]) E.push_back(e);
        sort(E.begin(), E.end(), [](const edge<T> &e1, const edge<T> &e2){return e1.cost < e2.cost;});
        UF uf(n);
        T res = 0;
        for(edge<T> &e: E){
            if(!uf.same(e.from, e.to)){
                uf.unite(e.from, e.to);
                res += e.cost;
            }
        }
        return res;
    }
    
    vector<int> toposo(){
        vector<int> ret(n, -1), in(n, 0);
        rep(i, n)for(edge<T> &e: G[i]) ++in[e.to];
        int cur = 0;
        stack<int> st;
        rep(i, n)if(!in[i]) st.push(i);
        while(!st.empty()){
            int v = st.top(); st.pop();
            ret[cur++] = v;
            for(edge<T> &e: G[v]){
                --in[e.to];
                if(!in[e.to]) st.push(e.to);
            }
        }
        return ret;
    }
    
    void scc_dfs(int v, vector<bool> &used, vector<int> &vs){
        used[v] = true;
        for(edge<T> &e: G[v])if(!used[e.to]) scc_dfs(e.to, used, vs);
        vs.push_back(v);
    }
    
    void scc_rdfs(int v, int k, vector<int> &cmp, vector<bool> &used, vector<vector<int>> &rG){
        used[v] = true;
        cmp[v] = k;
        for(int nv: rG[v])if(!used[nv]) scc_rdfs(nv, k, cmp, used, rG);
    }
    
    pair<int, vector<int>> scc(){
        vector<vector<int>> rG(n);
        rep(i, n)for(edge<T> &e: G[i]) rG[e.to].push_back(i);
        vector<bool> used(n, false);
        vector<int> vs;
        vector<int> cmp(n);
        rep(i, n)if(!used[i]) scc_dfs(i, used, vs);
        fill(used.begin(), used.end(), false);
        int k = 0;
        rrep(i, n)if(!used[vs[i]]) scc_rdfs(vs[i], k++, cmp, used, rG);
        return make_pair(k, cmp);
    }
};

int main(){
    int n, m;
    scanf("%d%d", &n, &m);
    Graph<int> g(n+m);
    UF uf(n);
    map<int, int> mp;
    int cur = 0;
    rep(i, m){
        int p, q, c;
        scanf("%d%d%d", &p, &q, &c);
        if(mp.find(c) == mp.end()) mp[c] = ++cur;
        g.add_edge(--p, n-1+mp[c], 1);
        g.add_edge(n-1+mp[c], p, 1);
        g.add_edge(--q, n-1+mp[c], 1);
        g.add_edge(n-1+mp[c], q, 1);
        uf.unite(p, q);
    }
    if(!uf.same(0, n-1)){
        printf("-1\n");
        return 0;
    }
    auto d = g.dijkstra(0);
    printf("%d\n", d[n-1] / 2);
}