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>
//#include <math.h>
using namespace std;

#define INF 1.1e9
#define LINF 1.1e18
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define BIT(x,n) bitset<n>(x)
#define PI 3.14159265358979323846

typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<P,ll> PP;

//-----------------------------------------------------------------------------

int n,m;
unordered_map<ll,vector<P> > g;
unordered_map<ll,int> d;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin>>n>>m;
	REP(i,m) {
		ll p,q,c;cin>>p>>q>>c;
		p--,q--;
		ll p_change=p|c<<30,q_change=q|c<<30;
		g[p].pb(P(p_change,1));
		g[p_change].pb(P(p,1));
		g[p_change].pb(P(q_change,0));
		g[q_change].pb(P(p_change,0));
		g[q_change].pb(P(q,1));
		g[q].pb(P(q_change,1));
	}
	priority_queue<P,vector<P>,greater<P> > pq;
	pq.push(P(0,0));
	while(!pq.empty()) {
		P p=pq.top();pq.pop();
		ll v=p.se;
		if(d[v]==0&&v!=0) d[v]=INF;
		if(d[v]<p.fi) continue;
		for(auto e:g[v]) {
			if(d[e.fi]==0&&e.fi!=0) d[e.fi]=INF;
			if(d[e.fi]<=d[v]+e.se) continue;
			d[e.fi]=d[v]+e.se;
			pq.push(P(d[e.fi],e.fi));
		}
	}

	cout<<(d[(ll)n-1LL]==INF||d[(ll)n-1LL]==0?-1LL:d[(ll)n-1]/2LL)<<endl;

	return 0;
}