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>
using namespace std;
const int maxn = 1e6 + 7;
int n, m, cnt = 1, num, head[maxn], d[maxn], inq[maxn];
struct Node {int v, nxt, w;} G[maxn << 1];
void ins(int u, int v, int w) {G[cnt] = (Node) {v, head[u], w}; head[u] = cnt++;}
void spfa() {
	memset(d, 0x3f, sizeof d);
	queue<int> q;
	q.push(1); inq[1] = 1; d[1] = 0;
	while(!q.empty()) {
		int u = q.front(); q.pop(); inq[u] = 0;
		for (int i = head[u]; i; i = G[i].nxt) {
			int v = G[i].v; 
			if(d[v] > d[u] + G[i].w) {
				d[v] = d[u] + G[i].w;
				if(!inq[v]) {inq[v] = 1; q.push(v);}
			}
		}
	}
}
#define pii pair<int, int>
map<pii, int> p;
int get(pii a) {
	if(p.find(a) != p.end()) return p[a];
	else return p[a] = ++num;
}
int main() {
	scanf("%d%d", &n, &m); num = n;
	for (int i = 1, x, y, c; i <= m; ++i) {
		scanf("%d%d%d", &x, &y, &c);
		int r1 = get(make_pair(x, c)), r2 = get(make_pair(y, c));
		ins(x, r1, 1); ins(r1, x, 1);
		ins(y, r2, 1); ins(r2, y, 1);
		ins(r1, r2, 0); ins(r2, r1, 0);
	}
	spfa();
	if(d[n] > 2 * n) puts("-1");
	else printf("%d\n", d[n] / 2);
	return 0;
}