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;
char buf[1 << 15], *fs, *ft;
inline char getc() {
  return (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 15, stdin), fs == ft))
             ? 0
             : *fs++;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getc();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getc();
  }
  return x * f;
}
const int MAXN = 200010;
int n, len;
long long ans;
int lin[MAXN], e[MAXN << 1], ver[MAXN << 1], nex[MAXN << 1];
long long f[MAXN][5];
inline void add(int x, int y, int z) {
  ver[++len] = y;
  nex[len] = lin[x];
  lin[x] = len;
  e[len] = z;
}
inline void dp(int x, int father) {
  for (int i = lin[x]; i; i = nex[i]) {
    int tn = ver[i];
    if (tn == father) continue;
    dp(tn, x);
    if (!e[i]) {
      ans += (f[tn][0] + 1) * (2 + 2 * f[x][0] + f[x][1] + f[x][3]);
      ans += f[tn][3] * (1 + f[x][0]);
      ans += f[tn][1] * (f[x][0] + 1);
      f[x][0] += f[tn][0] + 1;
      f[x][3] += f[tn][3] + f[tn][1];
    } else {
      ans += (f[tn][1] + 1) * (2 + f[x][0] + 2 * f[x][1] + f[x][2]);
      ans += f[tn][2] * (1 + f[x][1]);
      ans += f[tn][0] * (f[x][1] + 1);
      f[x][1] += f[tn][1] + 1;
      f[x][2] += f[tn][2] + f[tn][0];
    }
  }
}
int main() {
  n = read();
  for (int i = 1; i < n; ++i) {
    int x, y, z;
    x = read();
    y = read();
    z = read();
    add(x, y, z);
    add(y, x, z);
  }
  dp(1, 0);
  printf("%lld\n", ans);
  return 0;
}
