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>
struct edge {
  int to, nxt, w;
} e[200010 << 1];
int head[200010], f[200010][2], g[200010][2], n, i, x, y, z, sz = 0;
long long ans = 0;
void addedge(int x, int y, int z) {
  e[++sz].to = y;
  e[sz].nxt = head[x];
  e[sz].w = z;
  head[x] = sz;
}
void dfs(int x, int y) {
  int i;
  f[x][0] = g[x][1] = 1;
  for (i = head[x]; i; i = e[i].nxt)
    if (e[i].to != y) {
      dfs(e[i].to, x);
      if (e[i].w) {
        ans += 1LL * (f[x][0] + f[x][1]) * g[e[i].to][1] +
               1LL * g[x][1] * (f[e[i].to][0] + f[e[i].to][1]);
        f[x][1] += f[e[i].to][0] + f[e[i].to][1];
        g[x][1] += g[e[i].to][1];
      } else {
        ans += 1LL * f[x][0] * (g[e[i].to][0] + g[e[i].to][1]) +
               1LL * (g[x][0] + g[x][1]) * f[e[i].to][0];
        f[x][0] += f[e[i].to][0];
        g[x][0] += g[e[i].to][0] + g[e[i].to][1];
      }
    }
}
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    addedge(x, y, z);
    addedge(y, x, z);
  }
  dfs(1, 0);
  printf("%I64d\n", ans);
  return 0;
}
