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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
const int N=500010;
int tt;
deque <int> Q;
map <int,int> w[N];
int head[N],to[N*3],nxt[N*3],s[N*3],dis[N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline void lnk(int x,int y,int z,int u) {
    to[++tt]=y,nxt[tt]=head[x],s[tt]=z,head[x]=tt;
    to[++tt]=x,nxt[tt]=head[y],s[tt]=u,head[y]=tt;
}
int main() {
    int n,m;
    cin>>n>>m;
    for(int i=1,cnt=n;i<=m;i++) {
	int x=gi(),y=gi(),z=gi(),&X=w[x][z],&Y=w[y][z];
	if(!X) X=++cnt;
	if(!Y) Y=++cnt;
	lnk(X,Y,0,0),lnk(X,x,0,1),lnk(Y,y,0,1);
    }
    memset(dis,127,sizeof(dis));
    Q.push_back(1),dis[1]=0;
    while(!Q.empty()) {
	int x=Q.front();
	Q.pop_front();
	for(int i=head[x];i;i=nxt[i]) {
	    int y=to[i];
	    if(dis[y]>dis[x]+s[i]) {
		dis[y]=dis[x]+s[i];
		Q.push_back(y);
	    }
	}
    }
    if(dis[n]>1<<30) dis[n]=-1;
    cout<<dis[n];
    return 0;
}