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 pb push_back
#define F first
#define S second
#define all(x) x.begin(), x.end()

using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;

const int Maxn = 3e5 + 10;
const int Maxk = 15;
const ll Inf = INT_MAX;

vector<pll> G[Maxn];
ll dis[Maxn];
struct edge {
	ll u, c, num;
};
bool CMP(edge a, edge b){
	return a.c < b.c;
}

vector<edge> H[Maxn];

deque<ll> deq;

int main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ll n, m;
	cin >> n >> m;
	ll u, v, c;
	edge x;
	for(int i = n + 1; i <= n + m; i++){
		cin >> u >> v >> c;
		x.num = i;
		x.u = u;
		x.c = c;
		H[v].pb(x);
	
		x.num = i;
		x.u = v;
		x.c = c;
		H[u].pb(x);
	
		G[u].pb({i, 1});
		G[i].pb({u, 0});
	
		G[v].pb({i, 1});
		G[i].pb({v, 0});
	}
	for(int i = 1; i <= n; i++){
		sort(all(H[i]), CMP);
		for(int j = 0; j + 1 < H[i].size(); j++){
			if(H[i][j].c == H[i][j + 1].c){
				G[H[i][j].num].pb({H[i][j + 1].num, 0});
				G[H[i][j + 1].num].pb({H[i][j].num, 0});
			}
		}
	}
	
	memset(dis, 31, sizeof dis);
	deq.pb(1);
	dis[1] = 0;
	ll fr;
	while(deq.size()){
		fr = deq.front();
		deq.pop_front();
		
		for(auto E : G[fr]){
			if(dis[E.F] > dis[fr] + E.S){
				dis[E.F] = dis[fr] + E.S;
				if(E.S == 0) deq.push_front(E.F);
				else deq.push_back(E.F);
			}
		}
	}
	
	
	cout << (dis[n] > Inf ? -1 : dis[n]) << '\n';
	return 0;
}