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;
int fa[2][200005], siz[2][200005];
int get_fa(int i, int u) { return fa[i][u] == u ? u : get_fa(i, fa[i][u]); }
void merge(int i, int x, int y) {
  x = get_fa(i, x);
  y = get_fa(i, y);
  if (siz[i][x] > siz[i][y]) swap(x, y);
  siz[i][y] += siz[i][x];
  fa[i][x] = y;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    fa[0][i] = fa[1][i] = i;
    siz[0][i] = siz[1][i] = 1;
  }
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    merge(w, u, v);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (fa[0][i] == i) ans += 1LL * (siz[0][i]) * (siz[0][i] - 1);
    if (fa[1][i] == i) ans += 1LL * (siz[1][i]) * (siz[1][i] - 1);
    ans += 1LL * (siz[0][get_fa(0, i)] - 1) * (siz[1][get_fa(1, i)] - 1);
  }
  printf("%lld\n", ans);
  return 0;
}
