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 + 101;
int n, m, k;
vector<pair<int, int>> G[N];
vector<int> nodes[10][10];
bool bad[10][10];
unsigned long long lol[10][10];
unsigned long long ran() {
  unsigned long long res = 0;
  for (int i = 0; i <= 50; ++i) res = res * 2 + rand() % 2;
  return res;
}
unsigned long long w[N], all;
int out;
void brute(int x, unsigned long long hasz) {
  if (x == k + 1) {
    out += all == hasz;
    return;
  }
  for (int i = 1; i <= x; ++i) brute(x + 1, hasz + lol[x][i]);
}
int main() {
  srand(2137);
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    G[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end()), w[i] = ran();
    all += w[i];
    int sz = (int)G[i].size();
    for (int j = 0; j <= sz - 1; ++j) {
      nodes[sz][j + 1].push_back(G[i][j].second);
    }
  }
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= k; ++j)
      for (auto x : nodes[i][j]) lol[i][j] += w[x];
  brute(1, 0);
  printf("%d\n", out);
  return 0;
}
