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 <stdio.h>
#include <string.h>
#include <set>
#include <queue>
#define N 100010
using namespace std;
int n,m,d[N];
int ne=0,h[N];struct edge{int to,w,x;}e[400010];
inline void ins(int s,int t,int w) {
	e[++ne].to=t;e[ne].w=w;e[ne].x=h[s];h[s]=ne;
}
struct stat {
	int n,d;
	stat(){}
	stat(int a,int b){n=a;d=b;}
	inline bool operator>(const stat& a) const {
		return d!=a.d?d>a.d:n<a.n;
	}
};
set<int> vis[N];
inline int dij(int s,int t) {
	priority_queue<stat,vector<stat>,greater<stat> > q;
	stat u;int v,oo,nd;
	memset(d,33,sizeof(d));oo=*d;
	d[s]=0;q.push(stat(s,0));
	while(!q.empty()) {
		u=q.top();q.pop();
		if(u.d>d[u.n]) continue;
		for(int i=h[u.n];i;i=e[i].x) {
			v=e[i].to;nd=u.d+(vis[u.n].count(e[i].w)?0:1);
			if(nd<d[v]) {
				d[v]=nd;q.push(stat(v,nd));
				vis[v].clear();
			}
			if(nd==d[v]) vis[v].insert(e[i].w);
		}
	}
	return oo==d[t]?-1:d[t];
}
int main() {
	int a,b,c;
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;++i) {
		scanf("%d%d%d",&a,&b,&c);
		ins(a,b,c);ins(b,a,c);
	}
	printf("%d",dij(1,n));
	return 0;
}