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 1000100
using namespace std;
typedef pair<int,int> pii;
int n,m,newn,len;
vector<pii> bel[N];
unordered_map<int,int> s[N];
int dis[N],cx[N];
int beg[N],va[N<<2],to[N<<2],nex[N<<2];
bool inq[N];
queue<int> q;
inline void Add(int a,int b,int c){
	nex[++len]=beg[a],beg[a]=len,to[len]=b,va[len]=c;
	nex[++len]=beg[b],beg[b]=len,to[len]=a,va[len]=c;
}
int main(){
	scanf("%d%d",&n,&m);
	newn=n;
	for(int i=0;i<m;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		bel[c].push_back(pii(a,b));
	}
	for(int i=1;i<=1000000;i++){
		int len=bel[i].size();
		if(!len) continue;
		for(int j=0;j<len;j++){
			if(s[i][bel[i][j].first]==0) s[i][bel[i][j].first]=++newn;
			if(s[i][bel[i][j].second]==0) s[i][bel[i][j].second]=++newn;
			int l=s[i][bel[i][j].first],r=s[i][bel[i][j].second];
			cx[l]=bel[i][j].first;
			cx[r]=bel[i][j].second;
			Add(l,r,0);
			Add(l,cx[l],1);
			Add(r,cx[r],1);
		}
	}
	memset(dis,-1,sizeof(dis));
	q.push(1);dis[1]=0;
	while(!q.empty()){
		int st=q.front();q.pop();
		for(int i=beg[st];i;i=nex[i]){
			if(dis[to[i]]>dis[st]+va[i]||dis[to[i]]==-1){
				dis[to[i]]=dis[st]+va[i];
				if(!inq[to[i]]){
					inq[to[i]]=1;
					q.push(to[i]);
				}
			}
		}
		inq[st]=0;
	}
	if(dis[n]<0) puts("-1");
	else printf("%d\n",dis[n]/2);
	return 0;
}