Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define MAXN 100005
#define MP make_pair
#define tomin(a,b) (a>(b)?a=(b):a) 
#define inf 2000000000
#define Dis first
#define No second
using namespace std;
int N,M;
struct Edge{
    int p,c,NXT;
};
Edge EDGE[MAXN<<2];
int tot,HEAD[MAXN];
void add_Edge(int a,int b,int c){
    EDGE[++tot]=(Edge){b,c,HEAD[a]};
    HEAD[a]=tot;
}
set<int> sub[MAXN];
typedef pair<int,int> pii;
priority_queue<pii> heap;
int dis[MAXN];
void Dij(){
    for(int i=2;i<=N;i++)
        dis[i]=inf;
    heap.push(MP(0,1));
    while(!heap.empty()){
        pii now=heap.top();
        heap.pop();
        if(-now.Dis>dis[now.No])continue;
        for(int v=HEAD[now.No];v;v=EDGE[v].NXT){
            Edge &nxt=EDGE[v];
            int cost=dis[now.No];
            if(sub[now.No].find(nxt.c)==sub[now.No].end())
                cost++;
            if(dis[nxt.p]>cost){
                dis[nxt.p]=cost;
                heap.push(MP(-cost,nxt.p));
                sub[nxt.p].clear();
            }
            if(dis[nxt.p]==cost)
                sub[nxt.p].insert(nxt.c);
        }
    }
}
int main(){
    scanf("%d%d",&N,&M);
    for(int i=1,p,q,c;i<=M;i++){
        scanf("%d%d%d",&p,&q,&c);
        add_Edge(p,q,c);add_Edge(q,p,c);
    }
    Dij();
    printf("%d\n",dis[N]==inf?-1:dis[N]);
    return 0;
}