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:
import heapq
v,e=map(int,input().split())
edge=[]
for i in range(e):
  s,t,w=map(int,input().split())
  edge.append([w,s,t])
heapq.heapify(edge)
class UnionFind:
  def __init__(self,n):
    self.n=n
    self.r=[i for i in range(n)]
    self.rank=[1 for i in range(n)]
    self.cost=0
  def find(self,x):
    if x==self.r[x]:
      return x
    else:
      self.r[x]=self.find(self.r[x])
      return self.r[x]
  def union(self,x,y,c):
    rx=self.find(x)
    ry=self.find(y)
    if rx==ry:
      return
    if self.rank[ry]>self.rank[rx]:
      rx,ry=ry,rx
    self.r[ry]=rx
    self.rank[rx]+=self.rank[ry]
    self.cost+=c
uf=UnionFind(v)
while edge:
  path=heapq.heappop(edge)
  if uf.find(path[1])==uf.find(path[2]):
    continue
  uf.union(path[1],path[2],path[0])
print(uf.cost)