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 = 2e5 + 7;
const unsigned long long h1 = 1e9 + 7;
int N, M, K, du[maxN] = {0};
vector<pair<int, int> > E[maxN];
int ans = 0;
vector<int> have_du[10];
unsigned long long x, save[10][10], need, Bas[maxN];
void dfs(int deep, int tim) {
  if (!deep) {
    if (x == need) ans += tim;
  } else {
    if (have_du[deep].empty())
      dfs(deep - 1, tim * deep);
    else
      for (int i = 1; i <= deep; i++) {
        x += save[deep][i];
        dfs(deep - 1, tim);
        x -= save[deep][i];
      }
  }
}
int main() {
  Bas[0] = 1;
  for (int i = 1; i < maxN; i++) Bas[i] = Bas[i - 1] * h1;
  scanf("%d%d%d", &N, &M, &K);
  for (int i = 1; i <= N; i++) need += Bas[i];
  for (int i = 1, u, v, w; i <= M; i++) {
    scanf("%d%d%d", &u, &v, &w);
    du[u]++;
    E[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= N; i++) sort(E[i].begin(), E[i].end());
  for (int i = 1; i <= N; i++) have_du[du[i]].push_back(i);
  for (int i = 1; i <= K; i++) {
    for (int j = 1; j <= i; j++) {
      for (int u : have_du[i]) save[i][j] += Bas[E[u][j - 1].second];
    }
  }
  x = 0;
  dfs(K, 1);
  printf("%d\n", ans);
  return 0;
}
