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 = 200000 + 5;
int n, fa[N][2], sz[N][2];
long long ans = 0;
int find(int x, int c) {
  if (fa[x][c] == x) return x;
  return fa[x][c] = find(fa[x][c], c);
}
inline void merge(int u, int v, int c) {
  u = find(u, c), v = find(v, c);
  if (u == v) return;
  if (sz[u][c] < sz[v][c]) swap(u, v);
  fa[v][c] = u, sz[u][c] += sz[v][c];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    fa[i][0] = fa[i][1] = i;
    sz[i][0] = sz[i][1] = 1;
  }
  for (int i = 1; i <= n - 1; i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    merge(u, v, c);
  }
  for (int i = 1; i <= n; i++) {
    if (find(i, 0) == i) ans += 1LL * sz[i][0] * (sz[i][0] - 1);
    if (find(i, 1) == i) ans += 1LL * sz[i][1] * (sz[i][1] - 1);
    ans += 1LL * (sz[find(i, 0)][0] - 1) * (sz[find(i, 1)][1] - 1);
  }
  printf("%lld\n", ans);
  return 0;
}
