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;
vector<pair<int, int> > AdjList[200005];
int perm[] = {0, 1, 2, 3, 4, 5, 6, 7, 8};
bool forbidden[10][10];
int taken[200005][10][10];
bool conflict[10][10][10][10];
int n, m, k;
int ans = 0;
vector<int> stack1;
void rec(int len) {
  if (len == k) {
    ans++;
    return;
  }
  for (int i = 0; i <= len; i++) {
    if (forbidden[len][i]) {
      continue;
    }
    bool possible = true;
    for (int j = 0; j < len; j++) {
      if (conflict[j][stack1[j]][len][i]) {
        possible = false;
        break;
      }
    }
    if (possible) {
      stack1.push_back(i);
      rec(len + 1);
      stack1.pop_back();
    }
  }
  return;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    AdjList[u].push_back(pair<int, int>(w, v));
  }
  for (int i = 1; i <= n; i++) {
    sort((AdjList[i]).begin(), (AdjList[i]).end());
  }
  memset(taken, false, sizeof(taken));
  memset(forbidden, false, sizeof(forbidden));
  memset(conflict, false, sizeof(conflict));
  for (int i = 1; i <= n; i++) {
    int k2 = (int)AdjList[i].size() - 1;
    for (int j = 0; j <= k2; j++) {
      int v = AdjList[i][j].second;
      taken[v][k2][j]++;
    }
  }
  for (int v = 1; v <= n; v++) {
    for (int k2 = 0; k2 < k; k2++) {
      for (int j = 0; j <= k2; j++) {
        if (taken[v][k2][j] >= 2) {
          forbidden[k2][j] = true;
        }
        if (taken[v][k2][j]) {
          for (int k3 = 0; k3 < k; k3++) {
            for (int j3 = 0; j3 <= k3; j3++) {
              if (taken[v][k3][j3]) {
                conflict[k2][j][k3][j3] = true;
                conflict[k3][j3][k2][j] = true;
              }
            }
          }
        }
      }
    }
  }
  rec(0);
  printf("%d", ans);
  return 0;
}
