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>
long long dp[400010 / 2][4];
long long go[400010], nex[400010], head[400010], v[400010];
int cnt = 0;
void ct(long long x, long long y, long long w) {
  nex[++cnt] = head[x];
  go[cnt] = y;
  v[cnt] = w;
  head[x] = cnt;
}
void dfs(int x, int fa) {
  for (int i = head[x]; i; i = nex[i]) {
    int now = go[i];
    int w = v[i];
    if (now == fa) continue;
    if (w == 0) {
      dp[x][0]++;
      dfs(now, x);
      dp[x][0] += dp[now][0];
      dp[x][1] += dp[now][3] + dp[now][1];
    } else {
      dp[x][3]++;
      dfs(now, x);
      dp[x][2] += dp[now][0] + dp[now][2];
      dp[x][3] += dp[now][3];
    }
  }
}
void dfs2(int x, int fa) {
  for (int i = head[x]; i; i = nex[i]) {
    int now = go[i];
    int w = v[i];
    if (now == fa) continue;
    if (w == 0) {
      dp[now][0] = dp[x][0];
      dp[now][1] = dp[x][1] + dp[x][3] - dp[now][3];
    } else {
      dp[now][2] = dp[x][2] + dp[x][0] - dp[now][0];
      dp[now][3] = dp[x][3];
    }
    dfs2(now, x);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    long long x, y, w;
    scanf("%lld %lld %lld", &x, &y, &w);
    ct(x, y, w);
    ct(y, x, w);
  }
  dfs(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = ans + dp[i][0] + dp[i][1] + dp[i][3];
  }
  printf("%lld\n", ans);
}
