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;
#define all(a)a.begin(),a.end()
using ll=long long;
const int INF = 1<<30;
const ll INFll =1LL<<62;
const int mod= int(1e9)+7;
using P = pair<int,int>;
using Pll = pair<ll,ll>;
using ld=long double;

struct edge{
    P to;
    int cost;
};

int main(){
    int n,m;cin >>n>>m;
    map<P,vector<edge>>g;
    set<P>used;
    deque<edge>que;
    que.emplace_back(edge{P(0,-1),0});
    for (int i = 0; i < m; ++i) {
        int p,q,c;cin >>p>>q>>c;
        p--;q--;
        g[P(p,c)].emplace_back((edge){P(q,c),0});
        g[P(q,c)].emplace_back((edge){P(p,c),0});
        g[P(p,c)].emplace_back((edge){P(q,-1),0});
        g[P(q,c)].emplace_back((edge){P(p,-1),0});
        g[P(p,-1)].emplace_back((edge){P(q,c),1});
        g[P(q,-1)].emplace_back((edge){P(p,c),1});
    }
    int ans=-1;
    while(que.size()){
        edge es=que.front();
        que.pop_front();
        if(used.find(es.to)!=used.end())continue;
        used.insert(es.to);
        if(es.to.first==n-1){
            ans=es.cost;
            break;
        }
        for(auto itr:g[es.to]){
            if(used.find(itr.to)!=used.end())continue;
            if(itr.cost==0)que.emplace_front((edge){itr.to,es.cost});
            else que.emplace_back((edge){itr.to,es.cost+1});
        }
    }
    cout <<ans <<endl;
}