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 <iostream>
#include <vector>
#include <queue>
#include <utility>

#define ll long long
#define P pair<ll,ll>
#define MAX 300010
#define INF 100000000000
using namespace std;

vector<ll> G[MAX],H[MAX];
vector<P> E[1000010];
ll dist[MAX],did[MAX];
bool used[MAX];
queue<P> que;
ll so = 100000,ans;

void bfs(int s){
	que.push({0,s});
	while(!que.empty()){
		P p = que.front();
		que.pop();
		ll x = p.first,y = p.second;
		if(used[y]) continue;
		used[y] = true; dist[y] = x;
		for(int i=0;i<G[y].size();i++){
			if(!used[G[y][i]]){
				que.push({x+1,G[y][i]});
			}
		}
	}
}

void dfs(ll cur, ll p){
	if(did[cur]==p) return;
	did[cur] = p;
	G[so].push_back(cur);
	G[cur].push_back(so);
	for(int i=0;i<H[cur].size();i++){
		dfs(H[cur][i],p);
	}
}

int main(){
	ll i,j,n,m;
	cin >> n >> m;
	for(i=0;i<m;i++){
		ll p,q,c;
		cin >> p >> q >> c;
		p--; q--;
		E[c].push_back({p,q});
	}
	for(i=0;i<MAX;i++){
		did[i] = -1;
		dist[i] = -2;
		used[i] = false;
	}
	
	for(i=0;i<1000010;i++){
		vector<ll> v;
		for(auto& e:E[i]){
			H[e.first].push_back(e.second);
			H[e.second].push_back(e.first);
			v.push_back(e.first);
			v.push_back(e.second);
		}
		
		for(auto& r:v){
			if(did[r]!=i){
				dfs(r,i);
				so++;
			}
		}
		for(auto& e:E[i]){
			H[e.first].clear();
			H[e.second].clear();
		}
	}
	
	bfs(0);
	ans = dist[n-1]/2;
	cout  << ans << endl;
}