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;
int n, m, k;
vector<int> q(9);
set<pair<int, int>> bad;
bool check[9];
int ans = 0;
void go(int p) {
  if (p == k) {
    ++ans;
    return;
  }
  if (check[p]) {
    for (int j = 0; j <= p; ++j) {
      if (bad.count(make_pair(q[j] + 9 * j, q[p] + 9 * p))) {
        return;
      }
    }
    go(p + 1);
    return;
  }
  for (int i = 0; i <= p; ++i) {
    q[p] = i;
    bool ok = 1;
    for (int j = 0; j <= p; ++j) {
      if (bad.count(make_pair(q[j] + 9 * j, q[p] + 9 * p))) {
        ok = 0;
        break;
      }
    }
    if (!ok) continue;
    go(p + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  vector<int> out[n];
  vector<pair<int, int>> in[n];
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    out[--u].emplace_back(w);
    in[--v].emplace_back(u, w);
  }
  for (int i = 0; i < n; ++i) sort(out[i].begin(), out[i].end());
  for (int i = 0; i < n; ++i) {
    vector<int> t;
    for (auto &x : in[i]) {
      int u = x.first, w = x.second;
      t.push_back(
          (out[u].size() - 1) * 9 +
          (int)(lower_bound(out[u].begin(), out[u].end(), w) - out[u].begin()));
    }
    sort(t.begin(), t.end());
    if (t.size() == 0) {
      cout << 0;
      return 0;
    }
    if (t.size() == 1) {
      check[t[0] / 9] = 1;
      q[t[0] / 9] = (t[0] % 9);
    }
    for (int i = 1; i < t.size(); ++i) {
      if (t[i] == t[i - 1]) {
        bad.emplace(t[i], t[i]);
      }
    }
    auto it = unique(t.begin(), t.end());
    t.resize(distance(t.begin(), it));
    for (int i = 0; i < t.size(); ++i) {
      for (int j = i + 1; j < t.size(); ++j) {
        bad.emplace(t[i], t[j]);
      }
    }
  }
  go(0);
  cout << ans;
}
