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 = 2e5 + 50, K = 12;
int n, m, k, ans, tp, a[K];
bool ban[K][K][K][K], in[N][K][K], B[K][K];
vector<pair<int, int> > bban[K][K];
vector<pair<int, int> > v[N], v2[N];
pair<int, int> st[N];
void dfs(int x) {
  int tt = tp;
  if (x > k) {
    ans++;
    return;
  }
  for (int i = 1; i <= x; i++) {
    if (!B[x][i] && !ban[x][i][x][i]) {
      a[x] = i;
      for (int j = 0; j < bban[x][i].size(); j++) {
        int xx = bban[x][i][j].first, yy = bban[x][i][j].second;
        if (!B[xx][yy]) st[++tp] = make_pair(xx, yy), B[xx][yy] = 1;
      }
      dfs(x + 1);
      while (tp > tt) {
        int xx = st[tp].first, yy = st[tp].second;
        B[xx][yy] = 0;
        tp--;
      }
    }
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, x, y, z; i <= m; i++) {
    scanf("%d%d%d", &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 i = 1; i <= n; i++) {
    int sz = v[i].size();
    for (int j = 0; j < sz; j++) {
      int y = v[i][j].second, p = j + 1;
      if (in[y][sz][p]) {
        ban[sz][p][sz][p] = 1;
        continue;
      }
      for (int k = 0; k < v2[y].size(); k++) {
        int x = v2[y][k].first, z = v2[y][k].second;
        if (!ban[sz][p][x][z])
          ban[sz][p][x][z] = 1, bban[sz][p].push_back(make_pair(x, z));
        if (!ban[x][z][sz][p])
          ban[x][z][sz][p] = 1, bban[x][z].push_back(make_pair(sz, p));
      }
      v2[y].push_back(make_pair(sz, p)), in[y][sz][p] = 1;
    }
  }
  dfs(1);
  cout << ans;
  return 0;
}
