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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos((long double)-1.0);
const double EPS = 1e-10;
const int MOD = 1e9 + 7;
template <typename T>
void cmin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void cmax(T &x, T y) {
  if (y > x) x = y;
}
long long qpow(long long x, long long n, long long mod = MOD) {
  if (n < 0) return 0;
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long sq(long long x) { return x * x; }
long long read() {
  long long res = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) {
    res = res * 10 + (c - '0');
    c = getchar();
  }
  return res;
}
namespace Solver {
void InitOnce() { int t; }
int n, m, k;
vector<pair<int, int> > G[200005];
unsigned long long hc1[200005], hc2[200005];
unsigned long long To1[10][10], To2[10][10];
void Read() {
  int res = scanf("%d%d%d", &n, &m, &k);
  if (res == -1) exit(0);
  for (int i = 1; i <= n; ++i) G[i].clear();
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    G[u].emplace_back(w, v);
  }
  mt19937_64 rnd;
  rnd.seed(19981111);
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end());
    hc1[i] = rnd();
    hc2[i] = rnd();
  }
  for (int i = 0; i <= 9; ++i) {
    for (int j = 0; j <= 9; ++j) {
      To1[i][j] = 0;
      To2[i][j] = 0;
    }
  }
  for (int u = 1; u <= n; ++u) {
    int sz = G[u].size();
    for (int i = 0; i < sz; ++i) {
      int v = G[u][i].second;
      To1[sz][i + 1] ^= hc1[v];
      To2[sz][i + 1] ^= hc2[v];
    }
  }
}
unsigned long long okcode1, okcode2;
unsigned long long code1, code2;
int ans;
void dfs(int pos, int val) {
  code1 ^= To1[pos][val];
  code2 ^= To2[pos][val];
  if (pos == k)
    ans += (code1 == okcode1) && (code2 == okcode2);
  else {
    for (int i = 1; i <= pos + 1; ++i) dfs(pos + 1, i);
  }
  code1 ^= To1[pos][val];
  code2 ^= To2[pos][val];
  return;
}
void Solve() {
  code1 = 0;
  okcode1 = 0;
  code2 = 0;
  okcode2 = 0;
  for (int i = 1; i <= n; ++i) okcode1 ^= hc1[i];
  for (int i = 1; i <= n; ++i) okcode2 ^= hc2[i];
  ans = 0;
  dfs(1, 1);
  printf("%d\n", ans);
}
}  // namespace Solver
int main() {
  Solver::InitOnce();
  while (true) {
    Solver::Read();
    Solver::Solve();
  }
  return 0;
}
