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 long double pi = 3.141592653589793;
const int mod = 1e9 + 7;
const int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
int n, m, k, x, y, w, cwm1 = 1, cwm2 = 1;
vector<pair<int, int> > g[1 << 20];
int m1[15][15], m2[15][15];
int per[15], ans = 0;
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
uniform_int_distribution<int> uf(1, 1e9);
long long v[1 << 20];
void permute(int i) {
  if (i == k + 1) {
    int c1 = 1, c2 = 1;
    for (int i = 1; i <= k; i++) {
      c1 = (c1 * 1LL * m1[i][per[i]]) % mod1;
      c2 = (c2 * 1LL * m2[i][per[i]]) % mod2;
    }
    if (c1 == cwm1 && c2 == cwm2) {
      ans++;
    }
    return;
  }
  for (int j = 1; j <= i; j++) {
    per[i] = j;
    permute(i + 1);
  }
}
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    v[i] = uf(mt);
    cwm1 = (cwm1 * 1LL * v[i]) % mod1;
    cwm2 = (cwm2 * 1LL * v[i]) % mod2;
  }
  for (int i = 0; i < m; i++) {
    cin >> x >> y >> w;
    g[x].push_back({w, y});
  }
  for (int i = 1; i <= 10; i++) {
    for (int j = 1; j <= 10; j++) {
      m1[i][j] = m2[i][j] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    int oedge = g[i].size();
    for (int j = 0; j < oedge; j++) {
      m1[oedge][j + 1] = (m1[oedge][j + 1] * 1LL * v[g[i][j].second]) % mod1;
      m2[oedge][j + 1] = (m2[oedge][j + 1] * 1LL * v[g[i][j].second]) % mod2;
    }
  }
  permute(1);
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
