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 re() {
  int x;
  scanf("%d", &x);
  return x;
}
long long val[200020][10];
long long key[200020];
struct node {
  int to, w;
};
vector<node> ro[200020];
bool cmp(node l, node r) { return l.w < r.w; }
int n, m, k;
long long rt;
int ans;
int e[10];
void gao() {
  long long tt = 0;
  for (int i = 1; i <= k; i++) tt += val[i][e[i]];
  if (tt == rt) ans++;
}
void dfs(int p) {
  if (p > k) {
    gao();
    return;
  }
  for (int i = 1; i <= p; i++) e[p] = i, dfs(p + 1);
}
int main() {
  n = re();
  m = re();
  k = re();
  for (int i = 1; i <= n; i++) rt += (key[i] = rand());
  for (int i = 1; i <= m; i++) {
    int u = re(), v = re(), w = re();
    ro[u].push_back({v, w});
  }
  for (int i = 1; i <= n; i++) sort(ro[i].begin(), ro[i].end(), cmp);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < ro[i].size(); j++) {
      val[ro[i].size()][j + 1] += key[ro[i][j].to];
    }
  }
  dfs(1);
  cout << ans;
}
