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;
const long long llINF = 9223372036854775807;
const int INF = 2147483647;
const int maxn = 2e5 + 7;
const int maxm = 5e4 + 7;
const long long mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-12;
int n;
int fa[2][maxn];
int a[maxn], b[maxn];
int u[maxn], v[maxn], w[maxn];
long long ans;
void init() {
  for (int j = 0; j < 2; j++)
    for (int i = 0; i < maxn; i++) fa[j][i] = i;
}
int find(int x, int idx) {
  return x == fa[idx][x] ? x : fa[idx][x] = find(fa[idx][x], idx);
}
void merge(int x, int y, int idx) { fa[idx][find(x, idx)] = find(y, idx); }
int main(int argc, char const *argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) scanf("%d%d%d", &u[i], &v[i], &w[i]);
  init();
  for (int i = 1; i <= n - 1; i++) {
    if (w[i] == 0)
      merge(u[i], v[i], 0);
    else
      merge(u[i], v[i], 1);
  }
  for (int i = 1; i <= n; i++) a[find(i, 0)]++;
  for (int i = 1; i <= n; i++) b[find(i, 1)]++;
  for (int i = 1; i <= n; i++) ans += 1ll * a[find(i, 0)] * b[find(i, 1)];
  printf("%lld\n", ans - n);
  return 0;
}
