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 re {
  int x, w;
  const bool operator<(const re &a) const { return w < a.w; }
};
const int M = 1000000000 + 7;
const int N = 200000 + 10;
const int MAX_POSSIBILITES = 50;
int n, m, k, deg[N];
vector<re> e[N];
vector<re> possibilities;
int location[10][10];
int invalid[10][10];
int cntPossibilities;
long long sumBitMask[N];
long long bitmask[MAX_POSSIBILITES];
int cal(int x, long long mask) {
  if (x == k + 1) return 1;
  int res = 0;
  for (int i = (1); (i <= (x)); ++i) {
    if (!invalid[i][x] && (mask & (1LL << location[i][x]))) {
      res += cal(x + 1, (mask & bitmask[location[i][x]]));
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = (1); (i <= (m)); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    e[u].push_back({v, w});
  }
  for (int i = (1); (i <= (k)); ++i) {
    for (int j = (1); (j <= (i)); ++j) {
      possibilities.push_back({j, i});
      location[j][i] = (int)possibilities.size() - 1;
    }
  }
  cntPossibilities = possibilities.size();
  for (int i = (0); (i <= (cntPossibilities - 1)); ++i) {
    bitmask[i] = (1LL << cntPossibilities) - 1;
  }
  for (int i = (1); (i <= (n)); ++i) {
    deg[i] = e[i].size();
    sort(e[i].begin(), e[i].end());
    for (int j = (0); (j <= (deg[i] - 1)); ++j) {
      if (sumBitMask[e[i][j].x] & (1LL << location[j + 1][deg[i]])) {
        invalid[j + 1][deg[i]] = 1;
      }
      sumBitMask[e[i][j].x] |= (1LL << location[j + 1][deg[i]]);
    }
  }
  for (int i = (1); (i <= (n)); ++i) {
    for (int j = (0); (j <= (cntPossibilities - 1)); ++j)
      if (sumBitMask[i] & (1LL << j)) {
        bitmask[j] &= ~(sumBitMask[i] & ~(1LL << j));
      }
  }
  int res = cal(1, (1LL << cntPossibilities) - 1);
  cout << res;
}
