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> > g[200005];
bool tab[55][55];
vector<int> st, spis[15];
int color[200005], answer = 0;
void rec(int num, int k) {
  if (num == k + 1) {
    answer++;
    return;
  }
  for (int i = 1; i <= num; i++) {
    int nm = (num - 1) * num / 2 + i;
    bool fl = tab[nm][nm];
    for (int j = 0; j < (int)st.size(); j++)
      if (!tab[st[j]][nm]) fl = false;
    if (fl) {
      st.push_back(nm);
      rec(num + 1, k);
      st.pop_back();
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    spis[(int)g[i].size()].push_back(i);
  }
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) {
      int num1 = (i - 1) * i / 2 + j;
      for (int k1 = 0; k1 < (int)spis[i].size(); k1++)
        color[g[spis[i][k1]][j - 1].second]++;
      tab[num1][num1] = true;
      for (int k1 = 1; k1 <= n; k1++)
        if (color[k1] > 1) tab[num1][num1] = false;
      for (int z = i + 1; z <= k; z++)
        for (int w = 1; w <= z; w++) {
          int num2 = (z - 1) * z / 2 + w;
          for (int k2 = 0; k2 < (int)spis[z].size(); k2++)
            color[g[spis[z][k2]][w - 1].second]++;
          tab[num1][num2] = true;
          for (int k1 = 1; k1 <= n; k1++)
            if (color[k1] > 1) tab[num1][num2] = false;
          for (int k2 = 0; k2 < (int)spis[z].size(); k2++)
            color[g[spis[z][k2]][w - 1].second]--;
        }
      for (int k1 = 0; k1 < (int)spis[i].size(); k1++)
        color[g[spis[i][k1]][j - 1].second]--;
    }
  rec(1, k);
  cout << answer << '\n';
  return 0;
}
