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;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout<<#a<<" = "<<a<<endl
#define dd(a) cout<<#a<<" = "<<a<<" "
typedef long long ll;
typedef double db;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef pair<int, pii> iii;
//----
const int N=101010;
int dis[N];
bool vis[N];
vector<pii> G[N];
set<int> s[N];
int n, m;
int dij(){
	priority_queue<pii> q;
	q.push(mp(0, 0));
	memset(dis, 0x3f, sizeof(dis));
	while(sz(q)){
		pii c=q.top();q.pop();
		if(vis[c.se])continue;
		vis[c.se]=1;
		int d=-c.fi;
		rep(i, 0, sz(G[c.se])){
			int v=G[c.se][i].fi, cc=G[c.se][i].se, nd=d;
			if(!s[c.se].count(cc))nd++;
			if(nd>dis[v])continue;
			if(nd<dis[v]){
				dis[v]=nd;
				q.push(mp(-dis[v], v));
				s[v].clear();
			}
			s[v].insert(cc);
		}
	}
	return dis[n-1]==dis[n]?-1:dis[n-1];
}
int main(){
	scanf("%d%d", &n, &m);
	rep(i, 0, m){
		int u, v, c;
		scanf("%d%d%d", &u, &v, &c);
		u--, v--;
		G[u].pb(mp(v, c));
		G[v].pb(mp(u, c));
	}
	printf("%d\n", dij());
	return 0;
}
/*
3 3
1 2 1
2 3 1
3 1 2
1

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
2

2 0
-1
*/