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 maxm = 2e5 + 10;
vector<int> g[maxm];
pair<int, int> e[maxm];
int a[maxm], tot;
int sum[10][10];
int d[maxm];
int c[maxm];
int n, m, k;
int ans;
void check() {
  int temp = 0;
  for (int i = 1; i <= k; i++) {
    temp += sum[i][c[i]];
  }
  if (temp == tot) ans++;
}
void dfs(int x) {
  if (x > k) {
    check();
    return;
  }
  for (int i = 1; i <= x; i++) {
    c[x] = i;
    dfs(x + 1);
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    e[c] = {a, b};
  }
  for (int i = 1; i <= m; i++) {
    int a = e[i].first;
    int b = e[i].second;
    g[a].push_back(b);
    d[a]++;
  }
  for (int i = 1; i <= n; i++) {
    a[i] = rand();
    tot += a[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < d[i]; j++) {
      int v = g[i][j];
      sum[d[i]][j + 1] += a[v];
    }
  }
  dfs(1);
  cout << ans << endl;
  return 0;
}
