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;
#pragma warning(disable : 4996)
int N, A[1 << 18], B[1 << 18], C[1 << 18];
int col1[1 << 18], col2[1 << 18], cnt1[1 << 18], cnt2[1 << 18], cnts;
vector<int> G1[1 << 18], G2[1 << 18];
void dfs1(int pos) {
  if (col2[pos] >= 1) return;
  cnt2[cnts]++;
  col2[pos] = cnts;
  for (int i = 0; i < G2[pos].size(); i++) dfs1(G2[pos][i]);
}
void dfs2(int pos) {
  if (col1[pos] >= 1) return;
  cnt1[cnts] += cnt2[col2[pos]];
  col1[pos] = cnts;
  for (int i = 0; i < G1[pos].size(); i++) dfs2(G1[pos][i]);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N - 1; i++) {
    scanf("%d%d%d", &A[i], &B[i], &C[i]);
    if (C[i] == 0) {
      G1[A[i]].push_back(B[i]);
      G1[B[i]].push_back(A[i]);
    }
    if (C[i] == 1) {
      G2[A[i]].push_back(B[i]);
      G2[B[i]].push_back(A[i]);
    }
  }
  cnts = 0;
  for (int i = 1; i <= N; i++) {
    if (col2[i] >= 1) continue;
    cnts++;
    dfs1(i);
  }
  cnts = 0;
  for (int i = 1; i <= N; i++) {
    if (col1[i] >= 1) continue;
    cnts++;
    dfs2(i);
  }
  long long ans = 0;
  for (int i = 1; i <= N; i++) ans += cnt1[col1[i]];
  cout << ans - N << endl;
  return 0;
}
