Boboniu has a directed graph with n vertices and m edges.

The out-degree of each vertex is at most k.

Each edge has an integer weight between 1 and m. No two edges have equal weights.

Boboniu likes to walk on the graph with some specific rules, which is represented by a tuple (c_1,c_2,…,c_k). If he now stands on a vertex u with out-degree i, then he will go to the next vertex by the edge with the c_i-th (1≤ c_i≤ i) smallest weight among all edges outgoing from u.

Now Boboniu asks you to calculate the number of tuples (c_1,c_2,…,c_k) such that

  * 1≤ c_i≤ i for all i (1≤ i≤ k). 
  * Starting from any vertex u, it is possible to go back to u in finite time by walking on the graph under the described rules. 

Input

The first line contains three integers n, m and k (2≤ n≤ 2⋅ 10^5, 2≤ m≤ min(2⋅ 10^5,n(n-1) ), 1≤ k≤ 9).

Each of the next m lines contains three integers u, v and w (1≤ u,v≤ n,u≠ v,1≤ w≤ m), denoting an edge from u to v with weight w. It is guaranteed that there are no self-loops or multiple edges and each vertex has at least one edge starting from itself.

It is guaranteed that the out-degree of each vertex is at most k and no two edges have equal weight.

Output

Print one integer: the number of tuples.

Examples

Input


4 6 3
4 2 1
1 2 2
2 4 3
4 1 4
4 3 5
3 1 6


Output


2


Input


5 5 1
1 4 1
5 1 2
2 5 3
4 3 4
3 2 5


Output


1


Input


6 13 4
3 5 1
2 5 2
6 3 3
1 4 4
2 6 5
5 3 6
4 1 7
4 3 8
5 2 9
4 2 10
2 1 11
6 1 12
4 6 13


Output


1

Note

For the first example, there are two tuples: (1,1,3) and (1,2,3). The blue edges in the picture denote the c_i-th smallest edges for each vertex, which Boboniu chooses to go through.

<image>

For the third example, there's only one tuple: (1,2,2,2).

<image>

The out-degree of vertex u means the number of edges outgoing from u.
// upsolve with kaiboy
import java.io.*;
import java.util.*;

public class CF1394B extends PrintWriter {
	CF1394B() { super(System.out, true); }
	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		byte[] bb = new byte[1 << 15]; int i, n;
		byte getc() {
			if (i >= n) {
				try { n = in.read(bb); } catch (IOException e) { n = 0; }
				if (n <= 0) return 0;
				i = 0;
			}
			return bb[i++];
		}
		int nextInt() {
			byte c = 0; while (c <= ' ') c = getc();
			int a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1394B o = new CF1394B(); o.main(); o.flush();
	}

	int[] oo, oi; int[][] oho, ohi; Integer[] hh;
	void link(int[] oo, int[][] oh, int i, int h) {
		int o = oo[i];
		if (o >= 2 && (o & o - 1) == 0)
			oh[i] = Arrays.copyOf(oh[i], o << 1);
		oh[i][oo[i]++] = h;
	}
	int[] ww, aa, bb;
	boolean[][][][] bad; int[] aa_; int[][] dd; int cnt;
	void init(int n, int m, int k) {
		oo = new int[n]; oho = new int[n][2];
		oi = new int[n]; ohi = new int[n][2];
		hh = new Integer[n - 1];
		ww = new int[m]; aa = new int[m]; bb = new int[m];
		bad = new boolean[k][k][k][k]; aa_ = new int[k]; dd = new int[k][k];
	}
	void count(int k, int b) {
		if (b == k) {
			cnt++;
			return;
		}
out:
		for (int a = 0; a <= b; a++) {
			if (bad[b][a][b][a])
				continue;
			for (int b_ = 0; b_ < b; b_++) {
				int a_ = aa_[b_];
				if (bad[b_][a_][b][a])
					continue out;
			}
			aa_[b] = a;
			count(k, b + 1);
		}
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		init(n, m, k);
		for (int h = 0; h < m; h++) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			ww[h] = sc.nextInt();
			link(oo, oho, i, h);
			link(oi, ohi, j, h);
		}
		for (int i = 0; i < n; i++) {
			for (int o = 0; o < oo[i]; o++)
				hh[o] = oho[i][o];
			Arrays.sort(hh, 0, oo[i], (h1, h2) -> ww[h1] - ww[h2]);
			for (int o = 0; o < oo[i]; o++) {
				int h = hh[o];
				aa[h] = o;
				bb[h] = oo[i] - 1;
			}
		}
		for (int i = 0; i < n; i++) {
			for (int b = 0; b < k; b++)
				for (int a = 0; a <= b; a++)
					dd[b][a] = 0;
			for (int o = 0; o < oi[i]; o++) {
				int h = ohi[i][o], a = aa[h], b = bb[h];
				dd[b][a]++;
			}
			for (int b1 = 0; b1 < k; b1++)
				for (int a1 = 0; a1 <= b1; a1++) {
					if (dd[b1][a1] != 1) {
						if (dd[b1][a1] > 1)
							bad[b1][a1][b1][a1] = true;
						continue;
					}
					for (int b2 = b1 + 1; b2 < k; b2++)
						for (int a2 = 0; a2 <= b2; a2++) {
							if (dd[b2][a2] == 0)
								continue;
							bad[b1][a1][b2][a2] = true;
						}
				}
		}
		count(k, 0);
		println(cnt);
	}
}
