You are given a tree (an undirected connected acyclic graph) consisting of n vertices and n - 1 edges. A number is written on each edge, each number is either 0 (let's call such edges 0-edges) or 1 (those are 1-edges).

Let's call an ordered pair of vertices (x, y) (x ≠ y) valid if, while traversing the simple path from x to y, we never go through a 0-edge after going through a 1-edge. Your task is to calculate the number of valid pairs in the tree.

Input

The first line contains one integer n (2 ≤ n ≤ 200000) — the number of vertices in the tree.

Then n - 1 lines follow, each denoting an edge of the tree. Each edge is represented by three integers x_i, y_i and c_i (1 ≤ x_i, y_i ≤ n, 0 ≤ c_i ≤ 1, x_i ≠ y_i) — the vertices connected by this edge and the number written on it, respectively.

It is guaranteed that the given edges form a tree.

Output

Print one integer — the number of valid pairs of vertices.

Example

Input


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


Output


34

Note

The picture corresponding to the first example:

<image>
import sys;sys.setrecursionlimit(10**9)
class UnionFind:
  def __init__(self,n):
    self.n=[-1]*n
    self.r=[0]*n
    self.siz=n
  def find_root(self,x):
    if self.n[x]<0:
      return x
    else:
      self.n[x]=self.find_root(self.n[x])
      return self.n[x]
  def unite(self,x,y):
    x=self.find_root(x)
    y=self.find_root(y)
    if x==y:return
    elif self.r[x]>self.r[y]:
      self.n[x]+=self.n[y]
      self.n[y]=x
    else:
      self.n[y]+=self.n[x]
      self.n[x]=y
      if self.r[x]==self.r[y]:
        self.r[y]+=1
    self.siz-=1
  def root_same(self,x,y):
    return self.find_root(x)==self.find_root(y)
  def count(self,x):
    return -self.n[self.find_root(x)]
  def size(self):
    return self.siz

n=int(input())
ouf=UnionFind(n)
zuf=UnionFind(n)
for _ in range(n-1):
  a,b,c=map(int,input().split())
  a-=1
  b-=1
  if c==0:
    zuf.unite(a,b)
  else:
    ouf.unite(a,b)
ans=0
for i in range(n):
  m=zuf.count(i)
  if zuf.find_root(i)==i:ans+=m*(m-1)
  mm=ouf.count(i)
  if ouf.find_root(i)==i:ans+=mm*(mm-1)
  ans+=(m-1)*(mm-1)
print(ans)