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>
const int kN = 200000 + 5;
int n, m, k;
std::bitset<kN> in[10][10];
int cnt[10][10];
bool collision[10][10][10][10];
std::vector<std::pair<int, int>> edges[kN];
int selected[10];
bool ok(int i, int j) {
  for (int k = 1; k < i; ++k) {
    if (collision[i][j][k][selected[k]]) return false;
  }
  return true;
}
int search(int at, int num) {
  if (at == k + 1) {
    return num == n;
  }
  int ret = 0;
  for (int i = 1; i <= at; ++i) {
    if (ok(at, i)) {
      selected[at] = i;
      ret += search(at + 1, num + cnt[at][i]);
    }
  }
  return ret;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    --a;
    --b;
    edges[a].emplace_back(c, b);
  }
  for (int u = 0; u < n; ++u) {
    std::sort(edges[u].begin(), edges[u].end());
    int s = edges[u].size();
    for (int i = 0; i < s; ++i) {
      in[s][i + 1].set(edges[u][i].second);
    }
  }
  for (int i = 1; i <= 9; ++i)
    for (int j = 1; j <= i; ++j) cnt[i][j] = in[i][j].count();
  for (int i = 1; i <= 9; ++i) {
    for (int j = 1; j <= i; ++j) {
      for (int k = 1; k <= 9; ++k) {
        for (int r = 1; r <= k; ++r) {
          if ((in[i][j] & in[k][r]).count()) collision[i][j][k][r] = true;
        }
      }
    }
  }
  printf("%d\n", search(1, 0));
}
