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;
typedef pair<int,int> pii;
const int maxn=2e6+5;
unordered_map<int,int> head[maxn];
struct EDGE{
	int v,nex,c;
}edge[maxn];
int ecnt;
inline void add_edge(int u,int v,int c){
	edge[++ecnt]={v,head[u][c],c};head[u][c]=ecnt;
	edge[++ecnt]={u,head[v][c],c};head[v][c]=ecnt;
}
int main() {
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	int n,m;cin>>n>>m;
	bool vis[maxn],vvis[maxn];
	vector<vector<int>> nex(maxn);
	int vcnt=n;
	for(int i=1;i<=m;i++) {
		int u,v,c;cin>>u>>v>>c;
		add_edge(u,v,c);
	}
	std::function<void(int,int)> dfs= [&] (int u,int col) {
		nex[u].push_back(vcnt);
		nex[vcnt].push_back(u);
		vvis[u]=1;
		for(int i=head[u][col];i;i=edge[i].nex) if(!vis[i]){
			vis[i]=1;
			int v=edge[i].v;if(vvis[v]) continue;
			dfs(v,col);
		}
	};
	for(int i=1;i<=ecnt;i++) {
		if(!vis[i]) {
			++vcnt;
			dfs(edge[i].v,edge[i].c);
			for(auto x:nex[vcnt]) vvis[x]=0;
			// vis[i]=1;
		}
	}

	vector<int> dis(maxn);queue<int> Q;Q.push(1);dis[1]=1;
	while(!Q.empty()) {
		int u=Q.front();Q.pop();
		for(auto v:nex[u]) {
			if(dis[v]) continue;
			dis[v]=dis[u]+1;
			Q.push(v);
		}
	}
	cout<<(dis[n]?(dis[n]-1)/2:-1)<<endl;
    return 0;
}