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 mxn = 2e5 + 100;
int pa1[mxn], pa2[mxn], sz1[mxn], sz2[mxn];
int all1[mxn], all2[mxn];
int find(int *pa, int u) { return u == pa[u] ? u : pa[u] = find(pa, pa[u]); }
void merge(int *sz, int *pa, int u, int v) {
  if ((u = find(pa, u)) != (v = find(pa, v))) {
    if (sz[u] < sz[v]) swap(u, v);
    sz[u] += sz[v], pa[v] = u;
  }
}
int main() {
  int n, u, v, w;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    pa1[i] = pa2[i] = i;
    sz1[i] = sz2[i] = 1;
  }
  for (int i = 1; i < n; ++i) {
    scanf("%d%d%d", &u, &v, &w);
    if (w)
      merge(sz2, pa2, u, v);
    else
      merge(sz1, pa1, u, v);
  }
  for (int i = 1; i <= n; ++i) all1[find(pa1, i)]++;
  for (int i = 1; i <= n; ++i) all2[find(pa2, i)]++;
  long long ret = 0;
  for (int i = 1; i <= n; ++i) {
    ret += 1LL * all1[i] * (all1[i] - 1);
    ret += 1LL * all2[i] * (all2[i] - 1);
    u = all1[find(pa1, i)], v = all2[find(pa2, i)];
    ret += 1LL * (u - 1) * (v - 1);
  }
  printf("%lld\n", ret);
}
