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 N = 2e5 + 10;
int head[N], wth[N << 1], pnt[N << 1], nxt[N << 1], E;
long long dp[N][4], ans;
int n;
void add(int u, int v, int w) {
  E++;
  pnt[E] = v;
  nxt[E] = head[u];
  wth[E] = w;
  head[u] = E;
}
void dfs(int u, int f) {
  for (int i = head[u]; i; i = nxt[i]) {
    int v = pnt[i];
    if (v == f) continue;
    dfs(v, u);
    if (wth[i] == 1) {
      ans += (dp[u][1] + 1) * (dp[v][1] + 1) * 2ll;
      ans += (dp[u][1] + 1) * (dp[v][2] + dp[v][0]);
      ans += (dp[u][2] + dp[u][0]) * (dp[v][1] + 1);
      dp[u][1] += dp[v][1] + 1;
      dp[u][2] += dp[v][2] + dp[v][0];
    } else {
      ans += (dp[u][0] + 1) * (dp[v][0] + 1) * 2ll;
      ans += (dp[u][0] + 1) * (dp[v][3] + dp[v][1]);
      ans += (dp[u][3] + dp[u][1]) * (dp[v][0] + 1);
      dp[u][0] += dp[v][0] + 1;
      dp[u][3] += dp[v][3] + dp[v][1];
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    add(u, v, w);
    add(v, u, w);
  }
  dfs(1, 1);
  printf("%lld\n", ans);
}
