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:

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import java.io.*;
import java.util.*;

public class D {

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Solver solver = new Solver();
        solver.solve(in, out);
        out.close();
    }

    private static class Solver {

        // DSU
        private int[][] p;
        private int[][] sz;

        private void make(int n) {
            p = new int[n + 2][2];
            sz = new int[n + 2][2];
            for (int i = 1; i <= n; i++) {
                p[i][0] = p[i][1] = i;
                sz[i][0] = sz[i][1] = 1;
            }
        }

        private int find(int v, int c) {
            if (p[v][c] == v) return v;
            return p[v][c] = find(p[v][c], c);
        }

        private void unite(int v, int u, int c) {
            v = find(v, c);
            u = find(u, c);
            if (v != u) {
                if (sz[v][c] < sz[u][c]) {
                    v ^= u; u ^= v; v ^= u; // swap
                }
                p[u][c] = v;
                sz[v][c] += sz[u][c];
            }
        }
        //

        private void solve(InputReader in, PrintWriter out) {
            int n = in.nextInt();

            make(n);

            for (int i = 0; i < n - 1; i++) {
                int v = in.nextInt();
                int u = in.nextInt();
                int c = in.nextInt();

                unite(v, u, c);
            }

            long ans = 0;
            for (int i = 1; i <= n; i++) {
                if (p[i][0] == i)
                    ans += (long) sz[i][0] * (sz[i][0] - 1);
                if (p[i][1] == i)
                    ans += (long) sz[i][1] * (sz[i][1] - 1);
                ans += (long) (sz[find(i, 0)][0] - 1) * (sz[find(i, 1)][1] - 1);
            }

            out.println(ans);
        }
    }

    private static class InputReader {
        private BufferedReader br;
        private StringTokenizer st;

        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
        }

        private int nextInt() {
            return Integer.parseInt(next());
        }

        private long nextLong() {
            return Long.parseLong(next());
        }

        private double nextDouble() {
            return Double.parseDouble(next());
        }

        private String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }

        private String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
    }
}