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 reg register
typedef long long ll;
using namespace std;
const int MN=1e5+5;
const int MM=4e5+5;
int n,m,tot;
int to[MM*5],nxt[MM*5],h[MM+MN],cnt;
inline void ins(int s,int t){
	to[++cnt]=t;nxt[cnt]=h[s];h[s]=cnt;
	to[++cnt]=s;nxt[cnt]=h[t];h[t]=cnt;
}
int dis[MN+MM];
map<int,int>mp[MN];
typedef pair<int,int>pii;
priority_queue<pii,vector<pii>,greater<pii> >q;
int main(){
	scanf("%d%d",&n,&m);tot=n;
	for(reg int i=1,s,t,c;i<=m;i++){
		scanf("%d%d%d",&s,&t,&c);
		if(!mp[s][c])mp[s][c]=++tot,ins(mp[s][c],s);
		if(!mp[t][c])mp[t][c]=++tot,ins(mp[t][c],t);
		ins(mp[s][c],mp[t][c]);
	}
	memset(dis,0x3f,sizeof(dis));
	dis[1]=0;q.push(make_pair(0,1));
	while(!q.empty()){
		reg pii v=q.top();q.pop();
		if(v.first!=dis[v.second])continue;
		for(reg int i=h[v.second];i;i=nxt[i])
			if(dis[to[i]]>v.first+((v.second<=n)||(to[i]<=n)))
				dis[to[i]]=v.first+((v.second<=n)||(to[i]<=n)),
				q.push(make_pair(dis[to[i]],to[i]));
	}
	if(dis[n]==0x3f3f3f3f)puts("-1");
	else printf("%d\n",dis[n]/2);
	return 0;
}