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> v[200005][2];
long long dp[200005][2][2];
long long ans;
void dpf(int cur, int p) {
  for (int nxt : v[cur][0])
    if (nxt != p) {
      dpf(nxt, cur);
      dp[cur][0][0] += dp[nxt][0][0] + 1;
      dp[cur][0][1] += dp[nxt][0][1] + dp[nxt][1][1];
    }
  for (int nxt : v[cur][1])
    if (nxt != p) {
      dpf(nxt, cur);
      dp[cur][1][0] += dp[nxt][1][0] + dp[nxt][0][0];
      dp[cur][1][1] += dp[nxt][1][1] + 1;
    }
  long long ret = (dp[cur][0][0] + 2) * (dp[cur][1][1] + 2) - 4 +
                  dp[cur][1][0] + dp[cur][0][1];
  for (int nxt : v[cur][0])
    if (nxt != p) {
      ret += (dp[nxt][0][0] + 1) * (dp[cur][0][0] - dp[nxt][0][0] - 1);
      ret +=
          (dp[nxt][0][0] + 1) * (dp[cur][0][1] - dp[nxt][1][1] - dp[nxt][0][1]);
    }
  for (int nxt : v[cur][1])
    if (nxt != p) {
      ret += (dp[nxt][1][1] + 1) * (dp[cur][1][1] - dp[nxt][1][1] - 1);
      ret +=
          (dp[nxt][0][0] + dp[nxt][1][0]) * (dp[cur][1][1] - dp[nxt][1][1] - 1);
    }
  ans += ret;
}
int main() {
  int N;
  scanf("%d", &N);
  for (int i = 1; i < N; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    v[a][c].push_back(b);
    v[b][c].push_back(a);
  }
  dpf(1, 0);
  printf("%lld", ans);
}
