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>
namespace IO {
double start_time = 0.0;
void ct() {
  start_time = clock();
  return;
}
void fast_cin() {
  std::ios::sync_with_stdio(false);
  std::cin.tie();
}
void read_f(int flag = 0) {
  freopen("0.in", "r", stdin);
  if (!flag) freopen("0.out", "w", stdout);
}
void run_time() {
  std::cout << "\nESC in : " << (clock() - start_time) * 1000.0 / CLOCKS_PER_SEC
            << "ms" << std::endl;
}
}  // namespace IO
using namespace IO;
template <typename T>
bool bacmp(const T& a, const T& b) {
  return a > b;
}
template <typename T>
bool pecmp(const T& a, const T& b) {
  return a < b;
}
using namespace std;
const int N = 2e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const unsigned long long bs = 19491001;
unsigned long long _pow(unsigned long long a, unsigned long long b) {
  unsigned long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
unsigned long long st;
vector<pair<int, int> > g[N];
unsigned long long sum[16][16];
unsigned long long ans;
int n, m, k;
long long cnt;
void dfs(int x) {
  if (x > k) {
    cnt += (ans == st);
    return;
  }
  for (int i = 1; i <= x; i++) {
    ans += sum[x][i];
    dfs(x + 1);
    ans -= sum[x][i];
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) st = st * bs + 1;
  while (m--) {
    int x, y, w;
    scanf("%d%d%d", &x, &y, &w);
    g[x].push_back(make_pair(w, y));
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    for (int j = 0; j < g[i].size(); j++) {
      sum[g[i].size()][j + 1] += _pow(bs, g[i][j].second - 1);
    }
  }
  dfs(1);
  printf("%lld\n", cnt);
  return 0;
}
