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 maxn = 200100;
int n, m, k;
vector<pair<int, int> > adj[maxn];
bool touched[maxn][10][10];
bool mat[100][100];
int stk[10];
bool skip[10];
bool badchoice[100];
int ans = 0;
int mult = 1;
inline int f(int i, int j) { return i * 10 + j; }
void dfs(int k1) {
  if (k1 > k) {
    ans++;
    return;
  }
  if (skip[k1]) {
    dfs(k1 + 1);
    return;
  }
  for (int j = 0; j < k1; j++) {
    bool good = true;
    int x = f(k1, j);
    if (badchoice[x]) continue;
    for (int i = 1; i < k1; i++)
      if (mat[stk[i]][x]) {
        good = false;
        break;
      }
    if (good) {
      stk[k1] = x;
      dfs(k1 + 1);
    }
  }
}
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);
    adj[u].push_back(pair<int, int>(w, v));
  }
  for (int i = 1; i <= k; i++) skip[i] = true;
  for (int i = 1; i <= n; i++) {
    sort(adj[i].begin(), adj[i].end());
    int k1 = adj[i].size();
    skip[k1] = false;
    for (int j = 0; j < k1; j++) {
      if (touched[adj[i][j].second][k1][j]) badchoice[f(k1, j)] = true;
      touched[adj[i][j].second][k1][j] = true;
    }
  }
  for (int i = 1; i <= k; i++)
    if (skip[i]) mult *= i;
  for (int i = 1; i <= n; i++)
    for (int k1 = 1; k1 <= k; k1++)
      for (int j1 = 0; j1 < k1; j1++)
        if (touched[i][k1][j1])
          for (int k2 = 1; k2 <= k; k2++)
            for (int j2 = 0; j2 < k2; j2++)
              if (touched[i][k2][j2]) mat[f(k1, j1)][f(k2, j2)] = true;
  dfs(1);
  printf("%d\n", ans * mult);
}
