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[200005], ffa[200005];
int sz[200005], ssz[200005];
int f(int x) { return (fa[x] == x ? x : fa[x] = f(fa[x])); }
int ff(int x) { return (ffa[x] == x ? x : ffa[x] = ff(ffa[x])); }
struct _edge {
  int x, y, c;
} edge[200005];
int main() {
  long long ans = 0;
  for (int i = 1; i <= 200005 - 2; ++i) {
    fa[i] = ffa[i] = i;
    sz[i] = ssz[i] = 1;
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].c);
  }
  for (int i = 1; i < n; ++i) {
    if (edge[i].c == 1) {
      sz[f(edge[i].y)] += sz[f(edge[i].x)];
      fa[f(edge[i].x)] = f(edge[i].y);
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (f(i) == i) {
      ans += 1ll * sz[i] * (sz[i] - 1);
    }
  }
  for (int i = 1; i < n; ++i) {
    if (edge[i].c == 0) {
      ssz[ff(edge[i].y)] += ssz[ff(edge[i].x)];
      ffa[ff(edge[i].x)] = ff(edge[i].y);
    }
  }
  for (int i = 1; i <= n; ++i) {
    ans += 1ll * sz[f(i)] * (ssz[ff(i)] - 1);
  }
  cout << ans;
  return 0;
}
