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;
int n;
int tot;
int cost[200050 << 1];
int head[200050], edge[200050 << 1], nxt[200050 << 1];
void join(int u, int v, int w) {
  cost[tot] = w;
  edge[tot] = v;
  nxt[tot] = head[u];
  head[u] = tot++;
  cost[tot] = w;
  edge[tot] = u;
  nxt[tot] = head[v];
  head[v] = tot++;
}
void init() {
  tot = 0;
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    join(u, v, w);
  }
}
long long ans;
long long dp[200050][4];
void dfs(int i, int fa = 0) {
  long long tmp = 0;
  for (int k = head[i]; ~k; k = nxt[k]) {
    int j = edge[k], w = cost[k];
    if (fa == j) continue;
    dfs(j, i);
    long long t = 0;
    if (!w) {
      ++dp[j][0];
      t += dp[j][0] * 2 + dp[j][2] + dp[j][1];
      t += dp[i][0] * dp[j][0] * 2;
      t += dp[i][0] * (dp[j][2] + dp[j][1]) + (dp[i][1] + dp[i][2]) * dp[j][0];
      dp[i][0] += dp[j][0];
      dp[i][2] += dp[j][1] + dp[j][2];
    } else {
      ++dp[j][1];
      t += dp[j][1] * 2 + dp[j][3] + dp[j][0];
      t += dp[i][1] * dp[j][1] * 2;
      t += dp[i][1] * (dp[j][0] + dp[j][3]) + (dp[i][0] + dp[i][3]) * dp[j][1];
      dp[i][1] += dp[j][1];
      dp[i][3] += dp[j][0] + dp[j][3];
    }
    tmp += t;
  }
  ans += tmp;
}
void solve() {
  dfs(1);
  cout << ans << endl;
}
int main() {
  init();
  solve();
}
