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 mod 1000000007
#define sp ' '
#define intmax 2147483647
#define llmax 9223372036854775807
#define nyan "(=^･ω･^=)"
#define mkp make_pair
#define lP pair<ll, ll>
#define iP pair<int,int>
typedef long long ll;
using namespace std;

int N, M, K, A, B, C, D[100000], vis[100000];
map<int, vector<pair<int,bool>>>v[100000];
queue<int>q;


void dfs(int x, int a, int t) {
	vector<pair<int,bool>>&V = v[x][a];
	for (int i = 0; i < V.size(); ++i) {
		if (!V[i].second) {
			V[i].second = true;
			if (D[V[i].first] == t + 1 && vis[V[i].first] != a) {
				vis[V[i].first] = a;
				dfs(V[i].first, a, t);
			}
			if (D[V[i].first] == -1) {
				D[V[i].first] = t + 1;
				q.push(V[i].first);
				dfs(V[i].first, a, t);
			}
		}
	}
}

int main() {
	cin >> N >> M;
	while (M--) {
		cin >> A >> B >> C;
		--A;
		--B;
		v[A][C].push_back(mkp(B, false));
		v[B][C].push_back(mkp(A, false));
	}
	memset(D, -1, sizeof D);
	D[0] = 0;
	q.push(0);
	while (!q.empty()) {
		int x = q.front();
		q.pop();
		for (auto it = v[x].begin(); it != v[x].end(); ++it) {
			dfs(x, it->first, D[x]);
		}
	}
	cout << D[N - 1] << endl;
}