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 INF 0x7fffffff
#define ll long long
#define P pair<int,int>
#define MP(a,b) make_pair(a,b)
using namespace std;
int n,m,dis[200005],head,tail,t,h[200005];
priority_queue<P> q;
set<int> in[200005];
struct E{int v,c,next;}g[400005];
void ins(int u,int v,int c){g[++t].v=v,g[t].c=c,g[t].next=h[u],h[u]=t;}
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
    return x*f;
}
int main(){
	memset(dis,0x3f,sizeof dis);
	n=read(),m=read();
	for(int i=1;i<=m;++i){
		int u,v,c;
		u=read(),v=read(),c=read();
		ins(u,v,c); ins(v,u,c);
	}
	q.push(MP(0,1)); dis[1]=0;
	while(!q.empty()){
		int u=q.top().second,cost=-q.top().first;
		q.pop();
		if(cost>dis[u]) continue;
		for(int j=h[u];j;j=g[j].next){
			int v=g[j].v,c=g[j].c,d=dis[u]+(!in[u].count(c));
			if(d>dis[v]) continue;
			if(d<dis[v]){
				q.push(MP(-d,v));
				dis[v]=d;
				in[v].clear();
			}
			in[v].insert(c);
		}
	}
	printf("%d",dis[n]==1061109567?-1:dis[n]);
}