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 = 1000010;
const int inf = 0x3f3f3f3f;

int n, m, ids, tot;
map< pair<int, int>, int> H;

inline int getid(int u, int v)
{
	if(H.find(make_pair(u, v)) != H.end()) return H[make_pair(u, v)];
	return H[make_pair(u, v)] = ++ids;
}

struct Edge
{
	int to, nxt; int w;
}e[MAXN << 1];
int h[MAXN];

inline void add(int u, int v, int w)
{
	e[++tot] = (Edge){v, h[u], w}; h[u] = tot;
	e[++tot] = (Edge){u, h[v], w}; h[v] = tot;
}

struct Node
{
	int u; int d;
	bool operator < (const Node &x) const
	{
		return d > x.d;
	}
};
priority_queue<Node> q;
int dis[MAXN];

void dijkstra()
{
	for(int i = 1; i <= ids; ++i) dis[i] = inf;
	dis[1] = 0; q.push((Node){1, 0});
	while(!q.empty())
	{
		Node x = q.top(); q.pop();
		if(dis[x.u] != x.d) continue;
		for(int i = h[x.u]; i; i = e[i].nxt)
		{
			int y = e[i].to; int z = e[i].w;
			if(dis[y] > dis[x.u] + z)
			{
				dis[y] = dis[x.u] + z;
				q.push((Node){y, dis[y]});
			}
		}
	}
}

int main()
{
	scanf("%d %d", &n, &m); ids = n;
	for(int i = 1; i <= m; ++i)
	{
		int u, v, z; scanf("%d %d %d", &u, &v, &z);
		int a = getid(u, z), b = getid(v, z);
		add(u, a, 1), add(a, b, 0), add(v, b, 1);
	}
	dijkstra();
	printf("%d\n", dis[n] == inf ? -1 : dis[n] >> 1);
	return 0;
}