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;
inline long long getint() {
  long long num = 0, flag = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') flag = -1;
  while (c >= '0' && c <= '9') num = num * 10 + c - 48, c = getchar();
  return num * flag;
}
struct node {
  int u, v, w;
} e[400005];
int n;
int fir[400005], nxt[400005], to[400005], cnt;
int fa[400005], sz[2][400005];
long long ans;
inline int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int main() {
  n = getint();
  for (int i = 1; i < n; i++)
    e[i].u = getint(), e[i].v = getint(), e[i].w = getint();
  for (int k = 0; k <= 1; k++) {
    for (int i = 1; i <= n; i++) fa[i] = i, sz[k][i] = 1;
    memset(fir, 0, sizeof fir), cnt = 0;
    for (int i = 1; i < n; i++)
      if (e[i].w == k) {
        int r1 = find(e[i].u), r2 = find(e[i].v);
        fa[r2] = r1, sz[k][r1] += sz[k][r2];
      }
    for (int i = 1; i <= n; i++) sz[k][i] = sz[k][find(i)];
  }
  for (int i = 1; i <= n; i++) ans += 1ll * sz[0][i] * sz[1][i];
  printf("%I64d\n", ans - n);
}
