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 N = 1e5 + 10, M = 2e5 + 30;
int n, m;
struct edge {
	int y, nxt, id;
}e[M << 1];
int lin[M], len;
int dis[N], pos[N];
bool vis[N];
set<int> c[N];

inline int read() {
	int s = 0, f = 1;
	char ch;
	for(; ch < '0' || ch > '9'; ch = getchar())	if(ch == '-')	f = -1;
	for(; ch >= '0' && ch <= '9'; ch = getchar())	s = (s << 1) + (s << 3) + ch - '0';
	return s * f;
}

inline void add(int xx, int yy, int vv) {
	e[++len] = (edge){yy, lin[xx], vv};
	lin[xx] = len;
}

inline void Dijkstra() {
	priority_queue< pair < int, int > > q;
	memset(dis, 0x3f, sizeof(dis));
	memset(vis, 0, sizeof(vis));
	dis[1] = 0;
	q.push(make_pair(0, 1));
	while(!q.empty()) {
		int x = q.top().second; q.pop();
		if(vis[x])	continue;
		vis[x] = 1;
		for(int i = lin[x]; i; i = e[i].nxt) {
			int y = e[i].y, v = c[x].count(e[i].id) ? 0 : 1;
			if(dis[y] > dis[x] + v) {
				dis[y] = dis[x] + v;
				q.push(make_pair(-dis[y], y));
				c[y].clear();
				c[y].insert(e[i].id);
			} else if(dis[y] == dis[x] + v)	c[y].insert(e[i].id);
		}
	}
}

int main() {
	n = read(), m = read(); 
	if(m == 0) { puts("-1"); return 0; }
	for(int i = 1; i <= m; ++i) {
		int xx = read(), yy = read(), vv = read();
		add(xx, yy, vv);
		add(yy, xx, vv);
	}
	Dijkstra();
	printf("%d\n", dis[n] == 0x3f3f3f3f ? -1 : dis[n]);
	return 0;
}