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>
def addEdge(arr: [], x, y):
    x, y = getRoot(arr, x), getRoot(arr, y)
    if arr[x] <= arr[y]:
        arr[x] += arr[y]
        arr[y] = x
    else:
        arr[y] += arr[x]
        arr[x] = y


def getRoot(arr: [], id):
    while arr[id] > 0:
        id = arr[id]
    return id


def solve(zeroArr: [], oneArr: []):
    Ans = 0
    for i in range(1, len(zeroArr)):
        if zeroArr[i] < 0:
            Ans += zeroArr[i] * (-1) * (zeroArr[i] * (-1) - 1)
        if oneArr[i] < 0:
            Ans += oneArr[i] * (-1) * (oneArr[i] * (-1) - 1)
        if (zeroArr[i] != -1 and oneArr[i] != -1):
            one, zero = getRoot(oneArr, i), getRoot(zeroArr, i)
            Ans += (oneArr[one] * (-1) - 1) * (zeroArr[zero] * (-1) - 1)
    return Ans

n = int(input())
zeroArr, oneArr = [-1] * (n + 1), [-1] * (n + 1)
while n > 1:
    a, b, w = [int(x) for x in input().split()]
    if w == 0:
        addEdge(zeroArr, a, b)
    else:
        addEdge(oneArr, a, b)
    n -= 1
print(solve(zeroArr, oneArr))
