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<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#include<map>
#include<set>
#define MN 400005
#define inf 0x3f3f3f3f
using namespace std;
int n,m,num,head[MN],dis[MN];
struct edge{int to,next,w;}g[MN<<1];
void ins(int u,int v,int w){g[++num].next=head[u];head[u]=num;g[num].to=v;g[num].w=w;}
set<int> now[MN];
map<int,vector<int> > co[MN]; 
queue<int> q;
void dfs(int u,int col,int d){
	now[u].insert(col);if(dis[u]==inf)dis[u]=d,q.push(u);
	for(int i=0;i<co[u][col].size();i++)if(!now[co[u][col][i]].count(col))if(dis[co[u][col][i]]>=d)dfs(co[u][col][i],col,d);
}
int main(){
	scanf("%d%d",&n,&m);memset(dis,0x3f,sizeof(dis));int x,y,v;
	for(int i=1;i<=m;i++)scanf("%d%d%d",&x,&y,&v),ins(x,y,v),ins(y,x,v),co[x][v].push_back(y),co[y][v].push_back(x);q.push(1);dis[1]=0;
	while(!q.empty()){
		int tmp=q.front();q.pop();
		for(int i=head[tmp];i;i=g[i].next)if(!now[tmp].count(g[i].w))dfs(tmp,g[i].w,dis[tmp]+1);
	}printf("%d\n",dis[n]==inf?-1:dis[n]);
}