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 ii      pair<int,int>
#define pb      push_back
using namespace std;

set<int> check;
vector<vector<int>> gn;

void dfs(map<int,vector<int>> & g, int cur){
    check.insert(cur);
    gn[cur].pb(gn.size() - 1);
    gn.back().pb(cur);
    for(auto &nei:g[cur]) if(check.find(nei) == check.end()) dfs(g, nei);
}
int main() {
	int n,m; cin >> n >> m;
	vector<vector<ii>> comp(1e6);
	for(int i=0,a,b,c;i<m;i++){
	    cin >> a >> b >> c; --a,--b,--c;
	    comp[c].pb({a,b});
	}
	gn.resize(n);
	for(int i=0;i<1e6;i++){
	    map<int,vector<int>> g;
	    for(auto &e : comp[i]){
	        g[e.first].pb(e.second);
	        g[e.second].pb(e.first);
	    }
	    check.clear();
	    for(auto &v:g){
	        if(check.find(v.first) != check.end()) continue;
	        gn.resize(gn.size()+1);
	        dfs(g,v.first);
	    }
	}
	vector<int> dis(gn.size(), 1e9);
	std::queue<int> q;
	q.push(0); dis[0] = 0;
	for(int cur;q.size();q.pop()){
	    cur = q.front();
	    for(auto &nei:gn[cur]){
	        if(dis[nei] == 1e9) q.push(nei), dis[nei] = dis[cur] + 1;
	    }
	}
	cout << (dis[n-1] != 1e9 ? dis[n-1]/2 : -1) << endl;
	return 0;
}