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 DSU:

    # Disjoint Set Union (Union-Find) Data Structure
    def __init__(self, nodes):
        # Dictionary of parents
        self.p = {i: i for i in nodes}
        # Dictionary of ranks
        self.r = {i: 0 for i in nodes}
        # Dictionary of sizes
        self.s = {i: 1 for i in nodes}


    def nonrecurive_get(self, u):
        # Returns the identifier of the set that contains u, includes path compression
        v = u
        while self.p[v] != v:
            v = self.p[v]
        while self.p[u] != u:
            u, self.p[u] = self.p[u], v
        return u

    def get(self, u):
        # Recursive Returns the identifier of the set that contains u, includes path compression
        if u != self.p[u]:
            self.p[u] = self.get(self.p[u])
        return self.p[u]


    def union(self, u, v):
        # Unites the sets with identifiers u and v
        u = self.get(u)
        v = self.get(v)
        if u != v:
            if self.r[u] > self.r[v]:
                u, v = v, u
            self.p[u] = v
            if self.r[u] == self.r[v]:
                self.r[v] += 1
            self.s[v] += self.s[u]

    def size(self, u):
        u = self.get(u)
        return self.s[u]

def solve():
    n = int(input())
    nodes = [i + 1 for i in range(n)]
    dsus = [DSU(nodes), DSU(nodes)]
    for edge in range(n - 1):
        x, y, c = [int(s) for s in input().split(' ')]
        dsus[c].union(x, y)
    S = 0
    for v in range(1, n + 1):
        S += (dsus[0].size(v) * dsus[1].size(v) - 1)
    return S

print(solve())

