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
input = sys.stdin.readline


class UnionFind:
    def __init__(self, n):
        self.parent = [-1] * n
        self.cnt = n

    def root(self, x):
        if self.parent[x] < 0:
            return x
        else:
            self.parent[x] = self.root(self.parent[x])
            return self.parent[x]

    def merge(self, x, y):
        x = self.root(x)
        y = self.root(y)
        if x == y:
            return
        if self.parent[x] > self.parent[y]:
            x, y = y, x
        self.parent[x] += self.parent[y]
        self.parent[y] = x
        self.cnt -= 1
        
    def is_same(self, x, y):
        return self.root(x) == self.root(y)

    def get_size(self, x):
        return -self.parent[self.root(x)]
    
    def get_cnt(self):
        return self.cnt


n = int(input())
info = [list(map(int, input().split())) for i in range(n - 1)]

uf0 = UnionFind(n)
uf1 = UnionFind(n)
tree0 = [[] for i in range(n)]
tree1 = [[] for i in range(n)]
for i in range(n - 1):
    a, b, cost = info[i]
    a -= 1
    b -= 1
    if cost == 0:
        tree0[a].append(b)
        tree0[b].append(a)
        uf0.merge(a, b)
    else:
        tree1[a].append(b)
        tree1[b].append(a)
        uf1.merge(a, b)

ans0 = [0] * n
ans1 = [0] * n
for i in range(n):
    ans0[i] = uf0.get_size(i) - 1
    ans1[i] = uf1.get_size(i) - 1

ans = 0
for i in range(n):
    ans += ans0[i]
    ans += ans1[i]
    ans += ans0[i] * ans1[i]
print(ans)