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;

struct E{
	int x, y, d;
	bool operator<(const E &oth) const {
		return d > oth.d;
	}
};

int n, m;
map<int, vector<E>> e[200010];
map<int, int> md[200010];
priority_queue<E> pq;

void add(int ax, int ay, int bx, int by, int c){
	e[ax][ay].push_back({bx, by, c});
	e[bx][by].push_back({ax, ay, c});
}

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 0, x, y, c; i < m; i++){
		scanf("%d%d%d", &x, &y, &c);
		add(x, c, y, c, 0);
		add(x, c, x, 0, 1);
		add(y, c, y, 0, 1);
	}
	md[1][0] = 0;
	pq.push({1, 0, 0});
	while(!pq.empty()){
		E c = pq.top(); pq.pop();
		for(auto &i : e[c.x][c.y]){
			if(md[i.x].find(i.y) == md[i.x].end() || md[i.x][i.y] > c.d + i.d){
				md[i.x][i.y] = c.d + i.d;
				pq.push({i.x, i.y, c.d + i.d});
			}
		}
	}
	printf("%d\n", (md[n].find(0) == md[n].end()) ? (-1) : (md[n][0] / 2));
}