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;
long long N, M, K;
vector<pair<long long, long long> > outEdge[200100];
vector<pair<long long, long long> > inEdge[200100];
vector<long long> idOcc[20][20];
vector<bool> isTrue(200100, 0);
long long precomKali[20];
bool inValid[20][20][20][20];
long long num[20][20];
bool isValid(vector<pair<long long, long long> > &ed) {
  long long sum = 0;
  for (pair<long long, long long> u : ed) {
    for (pair<long long, long long> v : ed) {
      if (inValid[u.first][u.second][v.first][v.second]) {
        return false;
      }
    }
  }
  for (pair<long long, long long> u : ed) {
    sum += num[u.first][u.second];
  }
  return sum == N;
}
long long dfs(long long now, vector<pair<long long, long long> > &ed) {
  if (now == K + 1) {
    return isValid(ed) ? 1 : 0;
  }
  long long sum = 0;
  for (long long v = 1; v <= now; v++) {
    ed.push_back({now, v});
    sum += dfs(now + 1, ed);
    ed.pop_back();
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M >> K;
  long long u, v, w;
  for (long long i = 0; i < M; i++) {
    cin >> u >> v >> w;
    outEdge[u].push_back({w, v});
  }
  for (long long u = 1; u <= N; u++) {
    sort(outEdge[u].begin(), outEdge[u].end());
    for (long long j = 0; j < (long long)outEdge[u].size(); j++) {
      outEdge[u][j].first = j + 1;
      long long v = outEdge[u][j].second;
      inEdge[v].push_back({(long long)outEdge[u].size(), j + 1});
    }
  }
  for (long long u = 1; u <= N; u++) {
    for (pair<long long, long long> p : inEdge[u]) {
      idOcc[p.first][p.second].push_back(u);
      num[p.first][p.second]++;
    }
  }
  for (long long u = 1; u <= N; u++) {
    sort(inEdge[u].begin(), inEdge[u].end());
    long long sz = inEdge[u].size();
    for (long long i = 2; i < sz; i++) {
      if (inEdge[u][i] == inEdge[u][i - 2]) {
        inEdge[u][i - 2] = {-1, -1};
      }
    }
    vector<pair<long long, long long> > newV;
    for (pair<long long, long long> v : inEdge[u]) {
      if (v != pair<long long, long long>(-1, -1)) {
        newV.push_back(v);
      }
    }
    inEdge[u] = newV;
    sz = inEdge[u].size();
    for (int i = 0; i < sz; i++) {
      for (int j = 0; j < sz; j++) {
        if (i == j) continue;
        pair<long long, long long> x = inEdge[u][i];
        pair<long long, long long> y = inEdge[u][j];
        inValid[x.first][x.second][y.first][y.second] = 1;
        inValid[y.first][y.second][x.first][x.second] = 1;
      }
    }
  }
  vector<pair<long long, long long> > edges;
  long long ans = dfs(1, edges);
  cout << ans << endl;
  return 0;
}
