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 std::vector;
struct Edge {
  int u, v, w;
  int id;
  Edge() {}
  Edge(int u, int v, int w, int id) : u(u), v(v), w(w), id(id) {}
};
int type[200005];
int deg[200005];
vector<Edge> inEdge[200005], outEdge[200005];
int f[15][15];
int ban[55];
int keep[15];
long long pow2[55];
long long banMask[55];
int cnt[55];
int preWork(int k) {
  int total = 0;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      f[i][j] = ++total;
    }
  }
  pow2[0] = 1;
  for (int i = 1; i <= 50; i++) pow2[i] = pow2[i - 1] * 2ll;
  return total;
}
int ans = 0;
void dfs(int p, int n, long long have) {
  for (int i = 1; i <= p; i++) {
    int u = f[p][i];
    if (ban[u]) continue;
    if (have & banMask[u]) continue;
    if (p == n)
      ans++;
    else
      dfs(p + 1, n, have + pow2[u - 1]);
  }
}
int main() {
  int now = 1;
  for (int i = 2; i <= 9; i++) now *= i;
  int n, m, k;
  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);
    deg[u]++;
    outEdge[u].push_back(Edge(u, v, w, i));
    inEdge[v].push_back(Edge(u, v, w, i));
  }
  int total = preWork(k);
  for (int i = 1; i <= n; i++) {
    std::sort(outEdge[i].begin(), outEdge[i].end(),
              [&](Edge& u, Edge& v) { return u.w < v.w; });
    int outd = deg[i];
    int size = outEdge[i].size();
    for (int j = 0; j < size; j++) {
      int u = f[outd][j + 1];
      type[outEdge[i][j].id] = u;
    }
  }
  for (int i = 1; i <= total; i++) banMask[i] = 0;
  for (int i = 1; i <= n; i++) {
    long long have = 0;
    for (Edge edge : inEdge[i]) {
      int uid = type[edge.id];
      have |= pow2[uid - 1];
      cnt[uid]++;
    }
    for (Edge edge : inEdge[i]) {
      int uid = type[edge.id];
      if (cnt[uid] != 1)
        ban[uid] = 1;
      else {
        banMask[uid] |= (have ^ pow2[uid - 1]);
      }
    }
    for (Edge edge : inEdge[i]) {
      int uid = type[edge.id];
      cnt[uid]--;
    }
  }
  ans = 0;
  long long have = 0;
  dfs(1, k, have);
  printf("%d\n", ans);
  return 0;
}
