// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



typedef pair<int,int> pii;

typedef pair<int,pii> edge;



const int N = 1e5;



int n,m;

edge li[2*N+5];



vector<int> isi[2*N+5];

vector<int> punya[N+5];



int find_par(int x, int *par){

	if(par[x] == x) return x;

	return par[x] = find_par(par[x],par);

}



int par[N+5];

bool vis[N+5];

map<int,int> uda;

bool hdone[2*N+5];



int bfs(int x){

	memset(vis,0,sizeof(vis));

	memset(hdone,0,sizeof(hdone));

	queue<int> qx;

	while(!qx.empty()) qx.pop();

	qx.push(1);

	int ret = 0;

	while(!qx.empty()){

		ret++;

		int pan = qx.size();

		while(pan--){

			int cur = qx.front(); qx.pop();

			if(vis[cur]) continue;

			vis[cur] = 1;

			for(int i=0;i<punya[cur].size();i++){

				int head = punya[cur][i];

				if(hdone[head]) continue;

				hdone[head] = 1;

				for(int j=0;j<isi[head].size();j++){

					int now = isi[head][j];

					if(vis[now]) continue;

					if(now == n) return ret;

					qx.push(now);

				}

			}

		}

	}

	return -1;

}



int main(){

	scanf("%d%d",&n,&m);

	for(int i=1;i<=m;i++){

		int p,q,c;

		scanf("%d%d%d",&p,&q,&c);

		li[i] = make_pair(c,make_pair(p,q));

	}

	sort(li+1,li+m+1);

	int cnt = 0;

	for(int i=1;i<=n;i++) par[i] = i;

	int l = 1;

	for(int i=2;i<=m;i++){

		if(li[i].first == li[i-1].first) continue;

		for(int j=l;j<i;j++){

			int pu = find_par(li[j].second.first,par);

			int pv = find_par(li[j].second.second,par);

			if(pu == pv) continue;

			par[pu] = pv;

		}

		for(int j=l;j<i;j++){

			int u = li[j].second.first;

			int v = li[j].second.second;

			int pu = find_par(u,par);

			int pv = find_par(v,par);

			if(!vis[u]){

				if(uda.find(pu) == uda.end()){

					cnt++;

					uda[pu] = cnt;

				}

				int idx = uda[pu];

				punya[u].push_back(idx);

				isi[idx].push_back(u);

			}

			

			if(!vis[v]){

				if(uda.find(pv) == uda.end()){

					cnt++;

					uda[pv] = cnt;

				}

				int idx = uda[pv];

				punya[v].push_back(idx);

				isi[idx].push_back(v);

			}

		}

		for(int j=l;j<i;j++){

			int u = li[j].second.first;

			int v = li[j].second.second;

			par[u] = u;

			par[v] = v;

			vis[u] = 0;

			vis[v] = 0;

		}

		l = i;

		uda.clear();

	}

	for(int j=l;j<=m;j++){

		int pu = find_par(li[j].second.first,par);

		int pv = find_par(li[j].second.second,par);

		if(pu == pv) continue;

		par[pu] = pv;

	}

	for(int j=l;j<=m;j++){

		int u = li[j].second.first;

		int v = li[j].second.second;

		int pu = find_par(u,par);

		int pv = find_par(v,par);

		if(!vis[u]){

			if(uda.find(pu) == uda.end()){

				cnt++;

				uda[pu] = cnt;

			}

			int idx = uda[pu];

			punya[u].push_back(idx);

			isi[idx].push_back(u);

		}

		

		if(!vis[v]){

			if(uda.find(pv) == uda.end()){

				cnt++;

				uda[pv] = cnt;

			}

			int idx = uda[pv];

			punya[v].push_back(idx);

			isi[idx].push_back(v);

		}

	}

	for(int j=l;j<=m;j++){

		int u = li[j].second.first;

		int v = li[j].second.second;

		par[u] = u;

		par[v] = v;

		vis[u] = 0;

		vis[v] = 0;

	}

//	l = i;

	uda.clear();

	int ans = bfs(1);

	printf("%d\n",ans);

	return 0;

}

// Optimized solution to the same problem (faster):
/*    三玖天下第一！！！  */

#include<bits/stdc++.h>

using namespace std;

const int maxn=1e5,maxm=2e5;

int n,m,lin[maxn+5]={},len=0,dis[maxn+5];bool vis[maxn+5]={};set<int> rail[maxn+5];

struct edge{int y,next,rk;}e[maxm*2+5];

inline void insert(int xx,int yy,int rr){e[++len].next=lin[xx];lin[xx]=len;e[len].y=yy;e[len].rk=rr;} 

inline int dread()

{

	int x=0;bool f=1;char ch=getchar();

	while(ch<'0'||ch>'9'){if(ch=='-') f^=1;ch=getchar();}

	while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}

	return f?x:-x;

}

inline void in(){n=dread();m=dread();for(int i=1,a,b,c;i<=m;i++) a=dread(),b=dread(),c=dread(),insert(a,b,c),insert(b,a,c);}

inline void dijkstra()

{

	memset(dis,0x3f,sizeof(dis));priority_queue< pair<int,pair<int,int> > > q;q.push(make_pair(-(dis[1]=0),make_pair(1,0)));

	while(!q.empty())

	{

		pair<int,int> seat=q.top().second;q.pop();if(vis[seat.first]) continue;vis[seat.first]=1;

		for(int i=lin[seat.first],extra=!(rail[seat.first].count(e[i].rk));i;i=e[i].next,extra=!(rail[seat.first].count(e[i].rk)))

		{

			if(dis[e[i].y]>dis[seat.first]+extra) q.push(make_pair(-(dis[e[i].y]=dis[seat.first]+extra),make_pair(e[i].y,e[i].rk))),rail[e[i].y].clear(),rail[e[i].y].insert(e[i].rk);

			else if(dis[e[i].y]==dis[seat.first]+extra) rail[e[i].y].insert(e[i].rk);

		}

	}printf("%d\n",dis[n]==0x3f3f3f3f?-1:dis[n]);

}

int main(){in();dijkstra();return 0;}

// Fastest known solution for this problem: