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 pb push_back

const int mn=1e5+5;
const int mm=1e6+5;

int n,m;
int cnt;

struct edge{
    int to,dist;
};

vector<edge> g[mm];
int d[mm];
unordered_map<int,int> h[mn];

struct node{
    int id,dist;
    bool operator < (const node &p) const{
        return dist>p.dist;
    }
};

priority_queue<node> q;
void dijk(){
    memset(d,0x3c,sizeof d);
    q.push(node{1,d[1]=0});
    while(!q.empty()){
        node u=q.top();q.pop();
        if (d[u.id]<u.dist) continue;
        for(auto &e:g[u.id]){
            if (d[e.to]>u.dist+e.dist){
                q.push(node{e.to,d[e.to]=u.dist+e.dist});
            }
        }
    }
}

int main() {
    scanf("%d%d",&n,&m);
    cnt=n;
    for(int i=1;i<=m;i++){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        if (h[u].find(w)==h[u].end()){
            h[u][w]=++cnt;
            g[u].pb(edge{cnt,1});
            g[cnt].pb(edge{u,1});
        }
        u=h[u][w];
        if (h[v].find(w)==h[v].end()){
            h[v][w]=++cnt;
            g[v].pb(edge{cnt,1});
            g[cnt].pb(edge{v,1});
        };
        v=h[v][w];
        g[u].pb(edge{v,0});
        g[v].pb(edge{u,0});
    }
    dijk();
    if (d[n]>1e9) d[n]=-2;
    printf("%d\n",d[n]/2);
    return 0;
}