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;
int n;
vector<vector<pair<int, int>>> g;
vector<int> p;
vector<bool> was;
vector<int> sz;
long long res;
void calcSz(int v, int par = -1) {
  sz[v] = 1;
  for (int i = 0; i < (int)g[v].size(); ++i) {
    int to = g[v][i].first;
    if (to != par && !was[to]) {
      calcSz(to, v);
      sz[v] += sz[to];
    }
  }
}
int findCentroid(int v, int n, int par = -1) {
  for (int i = 0; i < (int)g[v].size(); ++i) {
    int to = g[v][i].first;
    if (to != par && !was[to] && 2 * sz[to] > n) {
      return findCentroid(to, n, v);
    }
  }
  return v;
}
int getInc(int v, int edge, int par = -1) {
  int res = 1;
  for (int i = 0; i < (int)g[v].size(); ++i) {
    int to = g[v][i].first;
    if (to != par && !was[to] && g[v][i].second >= edge) {
      res += getInc(to, g[v][i].second, v);
    }
  }
  return res;
}
int getDec(int v, int edge, int par = -1) {
  int res = 1;
  for (int i = 0; i < (int)g[v].size(); ++i) {
    int to = g[v][i].first;
    if (to != par && !was[to] && g[v][i].second <= edge) {
      res += getDec(to, g[v][i].second, v);
    }
  }
  return res;
}
int build(int v) {
  calcSz(v);
  v = findCentroid(v, sz[v]);
  was[v] = true;
  int inc0 = 0;
  int inc1 = 0;
  int dec0 = 0;
  int dec1 = 0;
  for (int i = 0; i < (int)g[v].size(); ++i) {
    int to = g[v][i].first;
    if (!was[to]) {
      int inc = getInc(to, g[v][i].second, v);
      int dec = getDec(to, g[v][i].second, v);
      if (g[v][i].second == 0) {
        res += inc * 1ll * (dec0 + 1);
        res += dec * 1ll * (inc0 + inc1 + 1);
        inc0 += inc;
        dec0 += dec;
      } else {
        res += inc * 1ll * (dec0 + dec1 + 1);
        res += dec * 1ll * (inc1 + 1);
        inc1 += inc;
        dec1 += dec;
      }
    }
  }
  for (int i = 0; i < (int)g[v].size(); ++i) {
    int to = g[v][i].first;
    if (!was[to]) {
      p[build(to)] = v;
    }
  }
  return v;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  g.resize(n);
  p.resize(n);
  was.resize(n);
  sz.resize(n);
  for (int i = 0; i < n - 1; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    --a;
    --b;
    g[a].push_back({b, c});
    g[b].push_back({a, c});
  }
  build(0);
  cout << res << endl;
  return 0;
}
