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 rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef pair<int, int>P;
 
struct st { int p, b, c; };
bool operator<(st a, st b) { return a.c>b.c; }
map<int, vector<int>>E[100000];
vector<int>G[100000];
map<int, int>d[100000];
int main() {
	int n, m; scanf("%d%d", &n, &m);
	rep(i, m) {
		int p, q, c; scanf("%d%d%d", &p, &q, &c); p--; q--;
		E[p][c].push_back(q); E[q][c].push_back(p);
		G[p].push_back(c); G[q].push_back(c);
	}
	priority_queue<st>que;
	d[0][0] = 0; que.push({ 0,0,0 });
	while (!que.empty()) {
		st p = que.top(); que.pop();
		if (d[p.p][p.b] != p.c)continue;
		if (d[p.p].find(0) == d[p.p].end() || d[p.p][0] > p.c) {
			d[p.p][0] = p.c;
			que.push({ p.p,0,p.c });
		}
		if (p.b == 0) {
			for (int v : G[p.p]) {
				if (d[p.p].find(v) == d[p.p].end() || d[p.p][v] > p.c + 1) {
					d[p.p][v] = p.c + 1;
					que.push({ p.p,v,p.c + 1 });
				}
			}
			continue;
		}
		for (int v : E[p.p][p.b]) {
			if (d[v].find(p.b) == d[v].end() || d[v][p.b] > p.c) {
				d[v][p.b] = p.c;
				que.push({ v,p.b,p.c });
			}
		}
	}
	int Min = INT_MAX;
	for (auto p : d[n - 1])Min = min(Min, p.second);
	if (Min == INT_MAX)puts("-1");
	else printf("%d\n", Min);
}