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>
using namespace std;
typedef long long LL;
const int INF=2147483647;
int N,M;
map<int,set<int> >ET[100000];
struct Pq
{
	int _u,_c,_cost;
	Pq(){}
	Pq(const int u,const int c,const int cost):_u(u),_c(c),_cost(cost){}
};
bool operator<(const Pq &a,const Pq &b){return a._cost>b._cost;}
int Solve()
{
	static bool vis[100000];
	for(int i=0;i<N;i++)vis[i]=false;
	priority_queue<Pq>pq;
	for(const auto &p:ET[0])for(const int nxt:p.second)
	{
		pq.push(Pq(nxt,p.first,1));
	}
	int ans=INF;
	while(!pq.empty())
	{
		const Pq q=pq.top();pq.pop();
		if(q._u==N-1)
		{
			if(q._cost<ans)ans=q._cost;
		}
		if(!vis[q._u])
		{
			vis[q._u]=true;
			for(const auto &p:ET[q._u])for(const int nxt:p.second)
			{
				pq.push(Pq(nxt,p.first,q._cost+1));
			}
		}
		for(const int nxt:ET[q._u][q._c])
		{
			pq.push(Pq(nxt,q._c,q._cost));
		}
		ET[q._u][q._c].clear();
	}
//	assert(0);
	return ans==INF?-1:ans;
}
int main()
{
//	freopen("in.txt","r",stdin);
	while(scanf("%d%d",&N,&M)==2)
	{
		for(int i=0;i<N;i++)ET[i].clear();
		for(int i=0,a,b,c;i<M;i++)
		{
			scanf("%d%d%d",&a,&b,&c);
			--a,--b;
			ET[a][c].insert(b),ET[b][c].insert(a);
		}
		printf("%d\n",Solve());
	}
	return 0;
}