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, sz[MAXN][2];
int f[MAXN][2];
inline int getfather(int x, int y) {
  return f[x][y] == x ? x : f[x][y] = getfather(f[x][y], y);
}
inline void merge(int x, int y, int z) {
  int xx = getfather(x, z);
  int yy = getfather(y, z);
  sz[yy][z] += sz[xx][z];
  f[xx][z] = yy;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) f[i][0] = f[i][1] = i, sz[i][0] = sz[i][1] = 1;
  for (int i = 1; i < n; ++i) {
    int x, y;
    x = read();
    y = read();
    merge(x, y, read());
  }
  for (int i = 1; i <= n; ++i) {
    int w1 = getfather(i, 0);
    int w2 = getfather(i, 1);
    if (w1 == i) ans += sz[w1][0] * (sz[w1][0] - 1);
    if (w2 == i) ans += sz[w2][1] * (sz[w2][1] - 1);
    ans += (sz[w1][0] - 1) * (sz[w2][1] - 1);
  }
  printf("%lld\n", ans);
  return 0;
}
