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>
n = int(input())

class UnionFind(object):
    def __init__(self, n):  # 初始化uf数组和组数目
        self.count = n
        self.uf = [i for i in range(n)]
        self.size = [1] * n # 每个联通分量的size

    def find(self, x):  # 判断节点所属于的组
        while x != self.uf[x]:
            self.uf[x] = self.uf[self.uf[x]]
            x = self.uf[x]
        return self.uf[x]

    def union(self, x, y):  # 连接两个节点
        x_root = self.find(x)
        y_root = self.find(y)
        if x_root == y_root:
            return
        if self.size[x_root] < self.size[y_root]:
            x_root, y_root = y_root, x_root
        self.uf[y_root] = x_root
        self.size[x_root] += self.size[y_root]
        self.size[y_root] = 0
        self.count -= 1


uf0 = UnionFind(n)
uf1 = UnionFind(n)

for i in range(n - 1):
    a, b, c = [int(j) for j in input().split(' ')]
    if c == 0:
        uf0.union(a-1, b-1)
    else:
        uf1.union(a-1, b-1)

res = 0


from collections import *


cnt0 = Counter(uf0.find(k) for k in range(n))
cnt1 = Counter(uf1.find(k) for k in range(n))

for c0, n0 in cnt0.items():
    res += n0 * (n0 - 1)
for c1, n1 in cnt1.items():
    res += n1 * (n1 - 1)

for i in range(n):
    n0 = cnt0[uf0.find(i)]
    n1 = cnt1[uf1.find(i)]
    res += (n0 - 1) * (n1 - 1)

print(res)
