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 void IO() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int mod = 1e9 + 7;
const int N = 3e5 + 5;
const int MAX = 0x7FFFFFFF;
const int MIN = 0x80000000;
const long long oo = 0x7FFFFFFFFFFFFFFF;
const long long sup = 0x8000000000000000;
int n;
int F_0[N];
int F_1[N];
long long sum_0[N] = {0};
long long sum_1[N] = {0};
int num_0[N];
int num_1[N];
struct node {
  int u, v, w;
} a[N];
int fa_0(int x) { return F_0[x] ^ x ? F_0[x] = fa_0(F_0[x]) : x; }
int fa_1(int x) { return F_1[x] ^ x ? F_1[x] = fa_1(F_1[x]) : x; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i <= n; i++) F_0[i] = F_1[i] = i, num_0[i] = num_1[i] = 1;
  for (int i = 1; i < n; i++) scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].w);
  long long res = 0;
  for (int i = 1; i < n; i++) {
    if (a[i].w) {
      int fx = fa_1(a[i].u), fy = fa_1(a[i].v);
      sum_1[fy] += sum_1[fx] + 1ll * num_1[fx] * num_1[fy];
      num_1[fy] += num_1[fx];
      F_1[fx] = fy;
    } else {
      int fx = fa_0(a[i].u), fy = fa_0(a[i].v);
      sum_0[fy] += sum_0[fx] + 1ll * num_0[fx] * num_0[fy];
      num_0[fy] += num_0[fx];
      F_0[fx] = fy;
    }
  }
  for (int i = 1; i <= n; i++) {
    int u0 = fa_0(i);
    int u1 = fa_1(i);
    res += 1ll * (num_0[u0] - 1) * (num_1[u1] - 1);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (F_1[i] == i) ans += sum_1[i];
    if (F_0[i] == i) ans += sum_0[i];
  }
  cout << (ans << 1) + res << endl;
}
