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.
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int n, m, k, d[N], c[N], A[20], ans = 0;
bool G[50][N], M[50][50], valid[50];
struct edge {
  int u, v, w;
} e[N];
bool cmp(edge a, edge b) { return a.w < b.w; }
int getid(int x, int y) { return x * (x - 1) / 2 + y; }
void dfs(int p) {
  if (p > k) {
    ++ans;
    return;
  }
  for (int i = 1; i <= p; ++i) {
    bool flag = true;
    int x = getid(p, i);
    if (!valid[x]) continue;
    for (int j = 1; j <= p - 1; ++j)
      if (!M[A[j]][x]) flag = false;
    if (flag) {
      A[p] = x;
      dfs(p + 1);
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) d[i] = c[i] = 0;
  for (int i = 1; i <= m; ++i)
    scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w), ++d[e[i].u];
  sort(e + 1, e + m + 1, cmp);
  memset(G, false, sizeof(G));
  memset(valid, true, sizeof(valid));
  for (int i = 1; i <= m; ++i) {
    ++c[e[i].u];
    if (G[getid(d[e[i].u], c[e[i].u])][e[i].v])
      valid[getid(d[e[i].u], c[e[i].u])] = false;
    G[getid(d[e[i].u], c[e[i].u])][e[i].v] = true;
  }
  memset(M, true, sizeof(M));
  int st = getid(k, k);
  for (int u = 1; u <= n; ++u)
    for (int i = 1; i <= st; ++i)
      if (G[i][u] && valid[i])
        for (int j = i + 1; j <= st; ++j)
          if (G[j][u]) M[i][j] = false;
  ans = 0;
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
