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;
vector<int> adj[200005];
map<pair<int, int>, int> cost;
long long int dp[200005][2], up[200005][2];
bool seen[200005];
void init() { return; }
inline long long int add(long long int a, long long int b) { return (a + b); }
inline long long int sub(long long int a, long long int b) { return (a - b); }
void dfs1(int src) {
  seen[src] = true;
  dp[src][0] = dp[src][1] = 0ll;
  for (auto it : adj[src]) {
    if (!seen[it]) {
      dfs1(it);
      if (cost[make_pair(src, it)] == 0) {
        dp[src][0] = add(dp[src][0], dp[it][0]);
        dp[src][0] = add(dp[src][0], 1ll);
      } else {
        dp[src][1] = add(dp[src][1], add(dp[it][0], dp[it][1]));
        dp[src][1] = add(dp[src][1], 1ll);
      }
    }
  }
  return;
}
void dfs2(int src) {
  seen[src] = true;
  if (src == 0) up[src][0] = up[src][1] = 0ll;
  for (auto it : adj[src]) {
    if (!seen[it]) {
      if (cost[make_pair(src, it)] == 0) {
        up[it][0] = add(up[src][0], sub(dp[src][0], dp[it][0]));
      } else {
        up[it][1] = add(
            dp[src][0],
            add(up[src][0],
                add(up[src][1], sub(dp[src][1], add(dp[it][0], dp[it][1])))));
      }
      dfs2(it);
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  int n;
  cin >> n;
  int x, y, c;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y >> c;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
    cost[make_pair(x, y)] = cost[make_pair(y, x)] = c;
  }
  memset(seen, false, sizeof(seen));
  dfs1(0);
  memset(seen, false, sizeof(seen));
  dfs2(0);
  long long int ans = 0ll;
  for (int i = 0; i < n; i++) {
    ans = add(ans, add(add(dp[i][0], dp[i][1]), add(up[i][0], up[i][1])));
  }
  cout << ans << endl;
  return 0;
}
