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 N = 200001;
const int K = 10;
int n, m, k, inter[K][K][K][K], invalid[K][K];
vector<pair<int, int> > v[N];
vector<int> g[K][K], h;
int ans;
void go(int idx) {
  if (idx == k + 1) {
    ans++;
    return;
  }
  h[idx] = 0;
  for (int i = 0; i < idx; i++) {
    int ok = 1;
    if (invalid[idx][h[idx]]) ok = 0;
    for (int j = 1; j < idx; j++) {
      if (inter[j][h[j]][idx][h[idx]]) {
        ok = 0;
      }
    }
    if (ok) go(idx + 1);
    h[idx]++;
  }
}
bool check(vector<int> &a, vector<int> &b) {
  for (auto &i : a) {
    if (binary_search(b.begin(), b.end(), i)) {
      return 1;
    }
  }
  return 0;
}
bool check(vector<int> &a) {
  for (int i = 1; i < (int)a.size(); i++) {
    if (a[i] == a[i - 1]) return 1;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    v[x].push_back(make_pair(z, y));
  }
  for (int i = 1; i <= n; i++) {
    sort(v[i].begin(), v[i].end());
    for (int j = 0; j < (int)v[i].size(); j++) {
      g[v[i].size()][j].push_back(v[i][j].second);
    }
  }
  for (int a = 1; a <= k; a++) {
    for (int c = 0; c < a; c++) {
      sort(g[a][c].begin(), g[a][c].end());
    }
  }
  for (int a = 1; a <= k; a++) {
    for (int c = 0; c < a; c++) {
      for (int b = a + 1; b <= k; b++) {
        for (int d = 0; d < b; d++) {
          inter[a][c][b][d] = check(g[a][c], g[b][d]);
        }
      }
    }
  }
  for (int a = 1; a <= k; a++) {
    for (int c = 0; c < a; c++) {
      invalid[a][c] = check(g[a][c]);
    }
  }
  h.resize(k + 1);
  go(1);
  cout << ans << endl;
}
