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 = 200005;
struct node {
  int to, val;
} a[maxn];
int n, u, v, w;
vector<node> vec[maxn];
int dp[maxn][2];
long long solve(int cur, int fa, int v) {
  long long res = dp[cur][0] + dp[cur][1], sum0 = 0, sum1 = 0;
  for (int i = 0; i < vec[cur].size(); i++) {
    auto sun = vec[cur][i];
    if (sun.to != fa) {
      if (sun.val)
        sum1 += dp[sun.to][0] + 1;
      else {
        sum0 += dp[sun.to][0] + 1;
        for (int j = 0; j < vec[sun.to].size(); j++) {
          auto son = vec[sun.to][j];
          if (son.to != cur && son.val) sum0 -= dp[son.to][0] + 1;
        }
      }
    }
  }
  for (int i = 0; i < vec[cur].size(); i++) {
    auto sun = vec[cur][i];
    if (sun.to != fa) {
      if (sun.val)
        dp[sun.to][1] = sum0 + sum1 - (dp[sun.to][0] + 1) + dp[cur][1] + 1;
      else {
        if (!v) dp[sun.to][1] += dp[cur][1];
        long long tot = dp[sun.to][0] + 1;
        for (int j = 0; j < vec[sun.to].size(); j++) {
          auto son = vec[sun.to][j];
          if (son.to != cur && son.val) tot -= dp[son.to][0] + 1;
        }
        dp[sun.to][1] += sum0 - tot + 1;
      }
      res += solve(sun.to, cur, sun.val);
    }
  }
  return res;
}
void dfs(int cur, int fa) {
  for (int i = 0; i < vec[cur].size(); i++) {
    if (vec[cur][i].to != fa) {
      dfs(vec[cur][i].to, cur);
    }
  }
  for (int i = 0; i < vec[cur].size(); i++) {
    auto sun = vec[cur][i];
    if (sun.to != fa) {
      if (sun.val)
        dp[cur][0] += dp[sun.to][0] + 1;
      else {
        dp[cur][0] += dp[sun.to][0] + 1;
        for (int j = 0; j < vec[sun.to].size(); j++) {
          auto grand = vec[sun.to][j];
          if (grand.to != cur && grand.val) dp[cur][0] -= dp[grand.to][0] + 1;
        }
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d %d %d", &u, &v, &w);
    vec[u].push_back(node{v, w});
    vec[v].push_back(node{u, w});
  }
  dfs(1, 0);
  printf("%lld\n", solve(1, 0, 0));
}
