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;
struct Triplet {
  int u, v, w;
  Triplet() {}
  Triplet(int u, int v, int w) : u(u), v(v), w(w) {}
};
bool cmp(const Triplet &T1, const Triplet &T2) { return T1.w < T2.w; }
const int MX = 10;
int degree[MX][MX][MX];
const int MX2 = 2e5 + 2;
vector<int> adj[MX2];
int cntRepeated[MX2][MX][MX];
int C[MX];
int k;
vector<int> adjF[MX];
bool vis[MX];
int cntVis = 0;
int indegree[MX];
int outdegree[MX];
bool forbidden[MX][MX][MX][MX];
int ans = 0;
void rec(int pos) {
  if (pos == k + 1) {
    cntVis = 0;
    for (int i = 0; i < (int)MX; i++) {
      adjF[i].clear();
      vis[i] = false;
      indegree[i] = outdegree[i] = 0;
    }
    for (int i = 1; i <= k; i++) {
      for (int j = i + 1; j <= k; j++) {
        if (degree[i][j][C[i]] > 0 || degree[j][i][C[j]] > 0) {
          adjF[i].push_back(j);
          adjF[j].push_back(i);
        }
      }
    }
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= k; j++) {
        if (i != j) {
          outdegree[i] += degree[i][j][C[i]];
          indegree[j] += degree[i][j][C[i]];
        }
      }
    }
    for (int i = 1; i <= k; i++) {
      for (int j = i + 1; j <= k; j++) {
        if (forbidden[i][C[i]][j][C[j]]) {
          return;
        }
      }
    }
    for (int i = 1; i <= k; i++) {
      if (indegree[i] != outdegree[i]) return;
    }
    ans++;
    return;
  }
  for (int i = 1; i <= pos; i++) {
    if (!forbidden[pos][i][pos][i]) {
      C[pos] = i;
      rec(pos + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m >> k;
  vector<Triplet> edges;
  for (int i = 0; i < (int)m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    edges.push_back(Triplet(u, v, w));
  }
  sort(edges.begin(), edges.end(), cmp);
  for (int i = 0; i < (int)m; i++) {
    adj[edges[i].u].push_back(edges[i].v);
  }
  for (int u = 0; u < (int)n; u++) {
    int fu = adj[u].size();
    int i = 1;
    for (int v : adj[u]) {
      int fv = adj[v].size();
      degree[fu][fv][i]++;
      cntRepeated[v][fu][i]++;
      i++;
    }
  }
  for (int u = 0; u < (int)n; u++) {
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= i; j++) {
        if (cntRepeated[u][i][j] > 1) {
          forbidden[i][j][i][j] = true;
        }
        for (int a = i + 1; a <= k; a++) {
          for (int b = 1; b <= a; b++) {
            if (cntRepeated[u][i][j] > 0 && cntRepeated[u][a][b] > 0) {
              forbidden[i][j][a][b] = forbidden[a][b][i][j] = true;
            }
          }
        }
      }
    }
  }
  rec(1);
  cout << ans << "\n";
  return 0;
}
