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;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;
int n, x, y, z;
int head[N], ver[N << 1], nxt[N << 1], edge[N << 1], tot;
long long ans, dp[N][4], num[4];
void addedge(int x, int y, int z) {
  ver[++tot] = y;
  edge[tot] = z;
  nxt[tot] = head[x];
  head[x] = tot;
}
void dfs(int x, int fa) {
  for (int i = head[x]; i != -1; i = nxt[i]) {
    int y = ver[i], z = edge[i];
    if (y == fa) continue;
    dfs(y, x);
    if (z == 0) {
      num[0] = dp[y][0] + 1;
      num[1] = 0;
      num[2] = dp[y][2] + dp[y][3];
      num[3] = 0;
    } else {
      num[0] = 0;
      num[1] = dp[y][0] + dp[y][1];
      num[2] = 0;
      num[3] = dp[y][3] + 1;
    }
    ans += dp[x][0] * (num[0] * 2 + num[2] + num[3]);
    ans += dp[x][1] * num[3];
    ans += dp[x][2] * num[0];
    ans += dp[x][3] * (num[3] * 2 + num[0] + num[1]);
    for (int i = 0; i < 4; i++) dp[x][i] += num[i];
  }
  ans += dp[x][0] * 2 + dp[x][3] * 2 + dp[x][1] + dp[x][2];
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  memset(head, -1, sizeof(head));
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> x >> y >> z;
    addedge(x, y, z);
    addedge(y, x, z);
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}
