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;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define per(i,a,n) for(int i=(n-1);i>=(a);i--)
#define fi first
#define se second
typedef pair <int,int> pII;
typedef long long ll;
const int INF = 0x3f3f3f3f;
//head
void read(int &x){
	char ch = getchar();x = 0;
	for (; ch < '0' || ch > '9'; ch = getchar());
	for (; ch >='0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
}
const int maxn = 1e5 + 10;
int n,m,ans,cnt;
int head[maxn],dis[maxn<<1];
struct Edge{
    int to,next,idx;
}G[maxn<<2];
void add(int u,int v,int id){
    G[cnt].to = v;
    G[cnt].next = head[u];
    G[cnt].idx = id;
    head[u] = cnt++;
}
deque<int> q;
void bfs(){
    for(int i = head[1];~i;i=G[i].next) dis[i/2] = 1,q.push_back(i/2);
    while(!q.empty()){
        int now = q.front();
        q.pop_front();
        for(int i = head[G[now*2].to];~i;i=G[i].next){
            if(G[i].idx == G[now*2].idx){
                if(dis[i/2] > dis[now]){
                    dis[i/2] = dis[now];q.push_front(i/2);
                }
            }else{
                if(dis[i/2] > dis[now] + 1){
                    dis[i/2] = dis[now]+1;q.push_back(i/2);
                }
            }
        }
        for(int i = head[G[now*2+1].to];~i;i=G[i].next){
            if(G[i].idx == G[now*2+1].idx){
                if(dis[i/2] > dis[now]){
                    dis[i/2] = dis[now];q.push_front(i/2);
                }
            }else{
                if(dis[i/2] > dis[now]+1){
                    dis[i/2] = dis[now]+1;q.push_back(i/2);
                }
            }
        }
    }
    ans = INF;
    for(int i = head[n];~i;i=G[i].next){
        if(ans > dis[i/2]) ans = dis[i/2];
    }
}
int main(){
#ifdef LOCAL
    freopen("a.in","r",stdin);
#endif
    while(~scanf("%d%d",&n,&m)){
        memset(head,-1,sizeof(int)*(n+2));cnt = 0;
        rep(i,0,m+2) dis[i] = INF;
        rep(i,0,m){
            int u,v,id;read(u);read(v);read(id);
            add(u,v,id);add(v,u,id);
        }
        bfs();
        printf("%d\n",ans==INF?-1:ans);
    }
    return 0;
}