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>
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n;
int fa[2][N], sz[2][N];
int Find(int x, int op) {
  return (x == fa[op][x] ? x : fa[op][x] = Find(fa[op][x], op));
}
void Union(int x, int y, int op) {
  int p = Find(x, op), q = Find(y, op);
  fa[op][p] = q;
  if (p != q) sz[op][q] += sz[op][p];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < N; i++) fa[0][i] = fa[1][i] = i, sz[0][i] = sz[1][i] = 1;
  for (int i = 1, u, v, w; i < n; i++) {
    scanf("%d%d%d", &u, &v, &w);
    Union(u, v, w);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    ans += 1LL * sz[0][Find(i, 0)] * sz[1][Find(i, 1)] - 1LL;
  printf("%lld\n", ans);
}
