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 = 5e5 + 5;
const int inf = 0x3f3f3f3f;

int n, m, head[maxn], cnt, dis[maxn], vis[maxn];
struct Graph{int to, nt, C;}e[maxn << 1];
set<int>S[maxn];
priority_queue<pair<int, int> >q;

inline int read() {
	char ch = getchar(); int u = 0, f = 1;
	while(!isdigit(ch)){if(ch == '-')f = -1; ch = getchar();}
	while(isdigit(ch)){u = u * 10 + ch - 48; ch = getchar();} return u * f;
}

inline void ini(int x, int y, int C) {e[++cnt] = (Graph){y, head[x], C}; head[x] = cnt;}

inline void Dij(int s) {
	memset(dis, 0x3f, sizeof(dis));
	memset(vis, 0, sizeof(vis));
	q.push(make_pair(dis[s] = 0, s));
	while(!q.empty()) {
		int now = q.top().second; q.pop();
		if(vis[now])continue; vis[now] = 1;
		for(register int i = head[now]; i; i = e[i].nt) {
			int v = e[i].to, w;
			w = S[now].count(e[i].C) ? 0 : 1;
			if(dis[v] > dis[now] + w) {
				dis[v] = dis[now] + w;
				q.push(make_pair(-dis[v], v));
				S[v].clear();
				S[v].insert(e[i].C);
			}
			else if(dis[v] == dis[now] + w) S[v].insert(e[i].C);
		}
	}
}

int main(){
	n = read(); m = read();
	for(register int i = 1; i <= m; i++){
		int x = read(), y = read(), C = read();
		ini(x, y, C); ini(y, x, C);
	}
	Dij(1);
	cout << (dis[n] == inf ? -1 : dis[n]);
	return 0;
}