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 MAXN = 2e5 + 7;
long long cnt0[MAXN], cnt1[MAXN];
int n, cnt;
bool mrk[MAXN][2];
vector<pair<int, int> > G[MAXN];
vector<int> nodes;
void dfs(int x, int peso) {
  mrk[x][peso] = true;
  cnt++;
  nodes.push_back(x);
  for (pair<int, int> edg : G[x]) {
    int viz = edg.first;
    int pi = edg.second;
    if (pi != peso || mrk[viz][peso]) continue;
    dfs(viz, peso);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int xi, yi, zi;
    scanf("%d %d %d", &xi, &yi, &zi);
    G[xi].push_back({yi, zi});
    G[yi].push_back({xi, zi});
  }
  for (int i = 1; i <= n; i++) {
    if (mrk[i][0]) continue;
    cnt = 0;
    dfs(i, 0);
    for (int node : nodes) {
      cnt0[node] = cnt;
    }
    nodes.clear();
  }
  for (int i = 1; i <= n; i++) {
    if (mrk[i][1]) continue;
    cnt = 0;
    dfs(i, 1);
    for (int node : nodes) {
      cnt1[node] = cnt;
    }
    nodes.clear();
  }
  long long ans = -n;
  for (int i = 1; i <= n; i++) {
    ans += cnt0[i] * cnt1[i];
  }
  cout << ans << endl;
}
