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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
using ll = long long;
const ll mod = 1e9 + 7;
const int maxn = 2e5 + 5;
int n, m, k;
vector<pair<int, int>> g[maxn];
vector<pair<int, int>> into[maxn];
int res;
int C[11];
bool bad[11][11][11][11];
bool collide[11][11];
bool ok(int d) {
  int c = C[d];
  for (int b = 1; b < d; b++) {
    int a = C[b];
    if (bad[a][b][c][d]) return false;
  }
  return true;
}
void print() {
  for (int i = 1; i <= k; i++) {
    cout << C[i] << ",";
  }
  cout << endl;
}
void go(int i) {
  if (i > k) {
    res++;
  } else {
    for (int j = 1; j <= i; j++) {
      if (collide[j][i]) {
        continue;
      }
      C[i] = j;
      if (ok(i)) {
        go(i + 1);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].push_back({w, v});
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
  }
  for (int i = 1; i <= n; i++) {
    int deg = g[i].size();
    int pos = 1;
    for (auto ed : g[i]) {
      int to = ed.second;
      into[to].push_back({pos++, deg});
    }
  }
  for (int i = 1; i <= n; i++) {
    if (into[i].empty()) {
      out(0);
    }
    vector<vector<bool>> have(11, vector<bool>(11, false));
    for (auto ed : into[i]) {
      int pos = ed.first;
      int deg = ed.second;
      if (have[pos][deg]) {
        collide[pos][deg] = true;
      }
      have[pos][deg] = true;
    }
    for (int a = 1; a <= k; a++) {
      for (int b = 1; b <= k; b++) {
        if (!have[a][b]) continue;
        for (int c = 1; c <= k; c++) {
          for (int d = 1; d <= k; d++) {
            if (have[c][d]) {
              bad[a][b][c][d] = true;
            }
          }
        }
      }
    }
  }
  go(1);
  cout << res << endl;
  return 0;
}
