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 mod = (1e9 + 7);
const long long inf = (1ll << 61);
const int MX = 3e5 + 9;
int n, x, y, w, ans[2][MX], vis[MX];
vector<int> v[MX], v1[MX], nodes;
long long ret = 0;
void dfs(int x, int p) {
  if (vis[x]) return;
  vis[x] = 1;
  ret++;
  nodes.push_back(x);
  for (auto pp : v[x]) {
    if (pp == p) continue;
    dfs(pp, x);
  }
}
void go(int x, int p) {
  if (vis[x]) return;
  vis[x] = 1;
  ret += ans[1][x] + 1;
  nodes.push_back(x);
  for (auto pp : v1[x]) {
    if (pp == p) continue;
    go(pp, x);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &x, &y, &w);
    if (w) {
      v[x].push_back(y);
      v[y].push_back(x);
    } else {
      v1[x].push_back(y);
      v1[y].push_back(x);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    ret = 0;
    nodes.clear();
    dfs(i, -1);
    puts("");
    for (auto pp : nodes) ans[1][pp] = ret - 1;
  }
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    ret = 0;
    nodes.clear();
    go(i, -1);
    for (auto pp : nodes) ans[0][pp] = ret - ans[1][pp] - 1;
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) res += ans[0][i] + ans[1][i];
  cout << res << endl;
}
