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 int maxn = 200010;
int a[maxn];
int f[2][maxn], sz[2][maxn];
int get(int pos, int x) {
  if (x == f[pos][x]) return x;
  return f[pos][x] = get(pos, f[pos][x]);
}
void merge(int pos, int x, int y) {
  int x1 = get(pos, x), y1 = get(pos, y);
  if (x1 == y1) return;
  f[pos][x1] = y1;
  sz[pos][y1] += sz[pos][x1];
}
int main() {
  int n, x, y, z;
  long long ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    f[0][i] = f[1][i] = i;
    sz[0][i] = sz[1][i] = 1;
  }
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    merge(z, x, y);
  }
  for (int i = 1; i <= n; i++) {
    ans += ((long long)sz[0][get(0, i)] * sz[1][get(1, i)]) - 1;
  }
  printf("%lld\n", ans);
}
