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>
#define X first
#define Y second
using namespace std;
typedef pair<int,int> pii;
const int N = 100005;

int n, m, ans, ts, s, vis[N];

queue<int> cur, nxt;
set<pii> adj[N];

void dfs (int C, int V) {
	if(!vis[C]) nxt.push(C);
	vis[C] = ts;
	while(true) {
		auto it = adj[C].lower_bound(pii(V,0));
		if(it == adj[C].end() || (it->X) != V) break;
		if(vis[it->Y] != ts && (!vis[it->Y] || vis[it->Y]>s)) dfs(it->Y, V);
		adj[C].erase(it);
	}
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++) {
		int A, B, C;
		scanf("%d%d%d",&A,&B,&C);
		adj[A].insert(pii(C, B));
		adj[B].insert(pii(C, A));
	}
	vis[1] = ++ts; cur.push(1);
	while(!vis[n] && !cur.empty()) {
		s = ts;
		while(!cur.empty()) {
			int C = cur.front(); cur.pop();
			while(!adj[C].empty()) {
				auto it = adj[C].begin();
				int A, B; tie(A, B) = (*it); ts++;
				if(!vis[B] || vis[B]>s) dfs(B, A);
				adj[C].erase(it);
			}
		}
		swap(cur, nxt); ans++;
	}
	printf("%d\n",(vis[n] ? ans : -1));
}