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;
struct ee {
  int to, last, val;
} e[400010];
int head[200010], g[200010], ng[200010], f[200010], nf[200010], tot, n;
long long ans;
void add(int u, int v, int w) {
  e[++tot].to = v;
  e[tot].val = w;
  e[tot].last = head[u];
  head[u] = tot;
}
void dfs1(int u, int fa) {
  for (int i = head[u]; i; i = e[i].last) {
    int to = e[i].to;
    if (to == fa) continue;
    dfs1(to, u);
    if (!e[i].val)
      g[u] += g[to] + 1;
    else
      f[u] += f[to] + g[to] + 1;
  }
}
void dfs2(int u, int fa) {
  ans += (long long)nf[u] + (long long)ng[u];
  for (int i = head[u]; i; i = e[i].last) {
    int to = e[i].to;
    if (to == fa) continue;
    if (!e[i].val)
      nf[to] = f[to], ng[to] = ng[u];
    else
      nf[to] = nf[u] + ng[u] - g[to], ng[to] = g[to];
    dfs2(to, u);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y, z; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    add(x, y, z);
    add(y, x, z);
  }
  dfs1(1, 0);
  nf[1] = f[1];
  ng[1] = g[1];
  dfs2(1, 0);
  printf("%lld", ans);
  return 0;
}
