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>
using namespace std;
typedef pair<int,int> ii;
const int N = 500005;
int n,m;
map<ii, int> id;
vector<ii> G[N];
int ct = 1;
priority_queue<ii, vector<ii>, greater<ii> > pq;
int dist[N];
int main(){
    scanf("%d%d",&n,&m);
    ct = n+1;
    for (int i = 0; i < m; i++){
        int p,q,c;
        scanf("%d%d%d",&p,&q,&c);
        int a,b;
        if (id.count({p,c})) a = id[ii(p,c)];
        else{
            a = id[ii(p,c)] = ct++;
            G[a].push_back({p,1});
            G[p].push_back({a,1});
        }
        if (id.count({q,c})) b = id[ii(q,c)];
        else{
            b = id[ii(q,c)] = ct++;
            G[b].push_back({q,1});
            G[q].push_back({b,1});
        }
        G[a].push_back({b,0});
        G[b].push_back({a,0});
    }
    memset(dist,-1,sizeof(dist));
    dist[1] = 0;
    pq.push({0,1});
    while (pq.size()){
        int d, u;
        tie(d,u) = pq.top(); pq.pop();
        if (d > dist[u]) continue;
        for (auto v : G[u]){
            int nd = d + v.second;
            if (dist[v.first] == -1 || dist[v.first] > nd){
                dist[v.first] = nd;
                pq.push({nd,v.first});
            }
        }
    }
    if (dist[n] == -1) printf("-1\n");
    else printf("%d",dist[n]/2);




}