Problem: Find the sum of weights of edges of the Minimum Spanning Tree for a given weighted undirected graph G = (V, E).

Constraints

* 1 ≤ |V| ≤ 10,000
* 0 ≤ |E| ≤ 100,000
* 0 ≤ wi ≤ 10,000
* The graph is connected
* There are no parallel edges
* There are no self-loops

Input


|V| |E|
s0 t0 w0
s1 t1 w1
:
s|E|-1 t|E|-1 w|E|-1


, where |V| is the number of vertices and |E| is the number of edges in the graph. The graph vertices are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target verticess of i-th edge (undirected) and wi represents the weight of the i-th edge.

Output

Print the sum of the weights of the Minimum Spanning Tree.

Examples

Input

4 6
0 1 2
1 2 1
2 3 1
3 0 1
0 2 3
1 3 5


Output

3


Input

6 9
0 1 1
0 2 3
1 2 1
1 3 7
2 4 1
1 4 3
3 4 1
3 5 1
4 5 6


Output

5

Solution (in PYTHON3):

Solution:
from collections import defaultdict 
import sys
import heapq
sys.setrecursionlimit(10**8)

def Find(x):	#xの根を返す
	global table

	if table[x] == x:
		return x
	else:
		table[x] = Find(table[x])	#親の更新(根を直接親にして参照距離を短く)
		return table[x]

def Union(x,y):	#xとyを繋げる
	x = Find(x)	
	y = Find(y)
	
	if x == y:
		return
	
	if rank[x] > rank[y]:
		table[y] = x
	else:
		table[x] = y
		if rank[x] == rank[y]:
			rank[y] += 1

def Check(x,y):
	if Find(x) == Find(y):
		return True
	else:
		return False



#N:頂点数 E:辺数 
N,E = map(int,input().split())

#Union-Find
table = [i for i in range(N)]	#木の親 table[x] == x なら根	
rank  = [1 for i in range(N)]	#木の長さ

# 優先度付きキューの作成
q = []
heapq.heapify(q)

#rout辞書
rout = defaultdict(set)
for i in range(E):
	a,b,w = list(map(int,input().split()))
	rout[a].add(b)
	rout[b].add(a)
	heapq.heappush(q,[w,a,b])

weight = 0
	
while q:
	w,a,b = heapq.heappop(q)
	
	if not Check(a,b):
		weight += w
		Union(a,b)
		
print(weight)