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], to[2 * N], nxt[2 * N], w[2 * N], tot;
void add(int x, int y, int z) {
  to[tot] = y, w[tot] = z, nxt[tot] = head[x], head[x] = tot++;
}
long long dp[N][2];
long long ans;
void dfs1(int u, int fa) {
  for (int i = head[u]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    int v = to[i];
    dfs1(v, u);
    if (w[i] == 1) {
      dp[u][1] += dp[v][0] + dp[v][1] + 1;
    } else {
      dp[u][0] += dp[v][0] + 1;
    }
  }
}
void dfs2(int u, int fa) {
  for (int i = head[u]; ~i; i = nxt[i]) {
    if (to[i] == fa) continue;
    int v = to[i];
    if (w[i] == 1) {
      dp[v][1] += (dp[u][1] - dp[v][1]) + (dp[u][0] - dp[v][0]);
    } else {
      dp[v][0] += dp[u][0] - dp[v][0];
    }
    dfs2(v, u);
  }
}
int main() {
  int n, x, y, z;
  scanf("%d", &n);
  for (int i = 0; i <= n; i++) head[i] = -1;
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += (dp[i][0] + dp[i][1]);
  printf("%lld\n", ans);
  return 0;
}
