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 N 100100
#define M 200100
using namespace std;
int a[M],b[M],c[M];
vector<int> g[M+N];
int dis[M+N];
int f[M];
int pre[1001000];
int find(int x){
	if(f[x]==x) return x;
	return f[x]=find(f[x]);
} 
void uni(int x,int y){
	int fx=find(x),fy=find(y);
	if(fx!=fy) f[fx]=fy;
}
int main(){
	int n,m;
	memset(dis,-1,sizeof(dis));
	memset(pre,-1,sizeof(pre));
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d%d%d",&a[i],&b[i],&c[i]);
		g[a[i]+m].push_back(i);
		g[b[i]+m].push_back(i);
		f[i]=i;
	}
	for(int i=m+1;i<=m+n;i++){
		for(int j=0;j<g[i].size();j++){
			int x=g[i][j];
			if(pre[c[x]]>=0) uni(pre[c[x]],x);
			pre[c[x]]=x;
		}
		for(int j=0;j<g[i].size();j++){
			int x=g[i][j];
			pre[c[x]]=-1;
		}
	}
	for(int i=0;i<m;i++){
		g[find(i)].push_back(a[i]+m);
		g[find(i)].push_back(b[i]+m);
	}
	for(int i=m+1;i<=m+n;i++){
		for(int j=0;j<g[i].size();j++){
			g[i][j]=find(g[i][j]);
		}
	}
	queue<int> q;
	q.push(m+1);
	dis[m+1]=0;
	while(!q.empty()){
		int u=q.front();
		q.pop();
		for(int i=0;i<g[u].size();i++){
			if(dis[g[u][i]]<0){
				dis[g[u][i]]=dis[u]+1;
				q.push(g[u][i]);
			}
		}
	}
	if(dis[m+n]<0) printf("%d\n",-1);
	else printf("%d\n",dis[m+n]/2);
	return 0;
}