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 ft first
#define sd second

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<pii, int> tri;

const int MAXN = 100 * 1000 + 123;



unordered_map<int, int> d[MAXN];
unordered_map<int, vector<int>> g[MAXN];
int n, m;

inline void inp();
inline void bfs();
inline bool upd(int&, const int);

int main() {
	inp();
	bfs();
	printf("-1");
	return 0;
}
inline void inp() {
	scanf("%d%d", &n, &m);
	int p, q, c;
	for (int i = 0; i < m; i++) {
		scanf("%d%d%d", &p, &q, &c);
		g[--p][c].push_back(--q), g[q][c].push_back(p);
	}
}
inline bool upd(int &x, const int y) { return ((x == 0 || y < x)? x = y, true: false); }
inline void bfs() {
	deque<tri> deq;
	d[0][0] = 0;
	deq.push_back({{0, 0}, 0});
	while (!deq.empty()) {
		tri v = deq.front();
		deq.pop_front();
		if (v.sd != d[v.ft.ft][abs(v.ft.sd)])
			continue;
		if (v.ft.ft == n - 1) {
			printf("%d", v.sd);
			exit(0);
		}
		if (v.ft.sd) {
			if (upd(d[v.ft.ft][0], v.sd))
				deq.push_front({{v.ft.ft, 0}, v.sd});
			for (auto &u : g[v.ft.ft][v.ft.sd]) {
				if (upd(d[u][v.ft.sd], v.sd))
						deq.push_front({{u, v.ft.sd}, v.sd});
			}
		}
		else {
			for (auto &i : g[v.ft.ft])
				if (upd(d[v.ft.ft][i.ft], v.sd + 1))
					deq.push_back({{v.ft.ft, i.ft}, v.sd + 1});
		}
	}
}