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>
#pragma GCC optimize(3)
using namespace std;
const int N = 200000;
int n, m, k;
vector<pair<int, int> > nei[N + 1];
vector<int> buc[10];
bitset<N + 1> bs[10][10];
bool ok[10][10][10][10];
bool fuck[10][10];
vector<int> now;
int ans = 0;
void dfs(int x = 1) {
  if (x == k + 1) {
    for (int i = 0; i < k; i++)
      if (fuck[i + 1][now[i]]) return;
    for (int i = 0; i < k; i++)
      for (int j = i + 1; j < k; j++)
        if (!ok[i + 1][now[i]][j + 1][now[j]]) return;
    ans++;
    return;
  }
  for (int i = 1; i <= x; i++) {
    now.push_back(i);
    dfs(x + 1);
    now.pop_back();
  }
}
int main() {
  cin >> n >> m >> k;
  while (m--) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    nei[x].push_back(make_pair(z, y));
  }
  for (int i = 1; i <= n; i++) sort(nei[i].begin(), nei[i].end());
  for (int i = 1; i <= n; i++) buc[nei[i].size()].push_back(i);
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < buc[i].size(); j++)
      for (int o = 1; o <= i; o++) {
        if (bs[i][o][nei[buc[i][j]][o - 1].second]) fuck[i][o] = true;
        bs[i][o].set(nei[buc[i][j]][o - 1].second);
      }
  }
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++)
      for (int o = i + 1; o <= k; o++)
        for (int p = 1; p <= o; p++)
          ok[i][j][o][p] =
              !fuck[i][j] && !fuck[o][p] && (bs[i][j] & bs[o][p]).none();
  dfs();
  cout << ans;
  return 0;
}
