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 << 23], *fc = buf, *tc = buf;
inline char gc() {
  if (fc == tc) {
    tc = (fc = buf) + fread(buf, 1, 1 << 23, stdin);
    if (fc == tc) return EOF;
  }
  return *fc++;
}
inline int read() {
  int x = 0, f = 1;
  char ch;
  while (!isdigit(ch = gc()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
  return f * x;
}
const int N = 200000 + 10;
int n;
int head[N], ver[N << 1], nxt[N << 1], edge[N << 1], tot;
inline void add(int x, int y, int z) {
  ver[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
  edge[tot] = z;
}
long long f[N][4];
long long ans;
void dfs(int x, int fa) {
  long long ret = 0;
  for (register int i = head[x]; i; i = nxt[i]) {
    int y = ver[i];
    if (y == fa) continue;
    dfs(y, x);
    if (edge[i]) {
      ret += f[x][1] * f[y][1] * 2;
      ret += f[x][1] * 2 + f[y][1] * 2 + 2;
      ret += f[x][0] * (f[y][1] + 1);
      ret += f[x][2] * (f[y][1] + 1);
      ret += f[y][0] * (f[x][1] + 1);
      ret += f[y][2] * (f[x][1] + 1);
      f[x][1] += f[y][1] + 1;
      f[x][2] += f[y][0] + f[y][2];
    } else {
      ret += f[x][0] * f[y][0] * 2;
      ret += f[x][0] * 2 + f[y][0] * 2 + 2;
      ret += (f[x][0] + 1) * f[y][1];
      ret += (f[y][0] + 1) * f[x][1];
      ret += (f[y][0] + 1) * f[x][3];
      ret += (f[x][0] + 1) * f[y][3];
      f[x][0] += f[y][0] + 1;
      f[x][3] += f[y][1] + f[y][3];
    }
  }
  ans += ret;
}
int main() {
  n = read();
  for (register int i = 1; i <= n - 1; ++i) {
    int x = read(), y = read(), z = read();
    add(x, y, z);
    add(y, x, z);
  }
  dfs(1, -1);
  cout << ans << endl;
  return 0;
}
