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>
template <typename Tp>
inline void read(Tp &x) {
  x = 0;
  bool f = true;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar()) f ^= ch == '-';
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + (ch ^ 48);
  x = f ? x : -x;
}
const int N = 2e5 + 7;
int Head[N], Cnt;
struct Edge {
  int to, nxt, w;
};
Edge e[N << 1];
inline void Add_edge(int x, int y, int w) {
  e[++Cnt] = (Edge){y, Head[x], w}, Head[x] = Cnt;
}
int F0[N], F1[N];
inline void Dfs1(int x, int last) {
  F1[x] = 1;
  for (int i = Head[x]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == last) continue;
    Dfs1(v, x);
    if (e[i].w)
      F1[x] += F1[v];
    else
      F0[x] += F1[v] + F0[v];
  }
}
inline void Dfs2(int x, int last) {
  for (int i = Head[x]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == last) continue;
    int a, b;
    if (e[i].w)
      a = F0[x], b = F1[x] - F1[v];
    else
      a = F0[x] - F1[v] - F0[v], b = F1[x];
    F0[v] += e[i].w ? 0 : a + b;
    F1[v] += e[i].w ? b : 0;
    Dfs2(v, x);
  }
}
int main() {
  int n;
  read(n);
  for (int i = 1; i < n; ++i) {
    int x, y, w;
    read(x), read(y), read(w);
    Add_edge(x, y, w), Add_edge(y, x, w);
  }
  Dfs1(1, 0);
  Dfs2(1, 0);
  long long Ans = 0;
  for (int i = 1; i <= n; ++i) Ans += F0[i] + F1[i] - 1;
  printf("%lld\n", Ans);
  return 0;
}
