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;
string impossible;
mt19937 mt_rand(
    chrono::high_resolution_clock::now().time_since_epoch().count());
const int N = 2e5 + 5;
const int L = 41;
int MOD = 1e9 + 7;
long long n, m, k, i, j;
bool f0, f1, f2;
const int MAXN = 2e3;
vector<vector<pair<int, int>>> g;
const int K = 10;
int ans = 0;
int c[K];
int in[N][K][K];
bool can[K][K], good[K][K][K][K];
void dfs(int v) {
  if (v == k) {
    ++ans;
    return;
  }
  c[v] = 0;
  while (c[v] <= v) {
    bool ok = can[v][c[v]];
    for (int i = 0; i < v; i++) {
      if (!good[i][c[i]][v][c[v]]) {
        ok = false;
        break;
      }
    }
    if (ok) dfs(v + 1);
    c[v]++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(11);
  int tests = 1;
  for (int te = 1; te <= tests; te++) {
    cin >> n >> m >> k;
    g = vector<vector<pair<int, int>>>(n);
    for (int i = 0; i < m; i++) {
      int u, v, w;
      cin >> u >> v >> w;
      u--;
      v--;
      g[u].push_back({w, v});
    }
    memset((in), (0), sizeof(in));
    ;
    for (int i = 0; i < n; i++) {
      sort(g[i].begin(), g[i].end());
      int d = g[i].size();
      for (int j = 0; j < d; j++) in[g[i][j].second][d - 1][j] += 1;
    }
    for (int i = 0; i < k; i++) {
      for (int j = 0; j <= i; j++) {
        can[i][j] = 1;
      }
    }
    for (int i = 0; i < k; i++) {
      for (int j = 0; j <= i; j++) {
        for (int p = 0; p < k; p++) {
          for (int q = 0; q <= p; q++) {
            good[i][j][p][q] = 1;
          }
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < k; j++) {
        for (int p = 0; p <= j; p++) {
          if (in[i][j][p] >= 2) can[j][p] = 0;
          if (in[i][j][p] > 0) {
            for (int q = j + 1; q < k; q++) {
              for (int r = 0; r <= q; r++) {
                if (in[i][q][r] > 0) good[j][p][q][r] = 0;
              }
            }
          }
        }
      }
    }
    dfs(0);
    cout << ans << "\n";
  }
  return 0;
}
