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 n0 = 2e5 + 123;
int n, m, k, cnt[10][10], bad[10][10], ans, num[10][10],
    badpair[10][10][10][10];
bitset<n0> groups[10][10];
vector<int> in[n0];
vector<array<int, 2>> out[n0];
array<int, 2> pt[n0];
int used[n0], arr[10];
void rec(int pos) {
  if (pos > 1)
    for (int i = 1; i < pos - 1; i++)
      if (badpair[i][arr[i]][pos - 1][arr[pos - 1]]) return;
  if (pos == k + 1) {
    ans++;
    return;
  }
  for (int i = 1; i <= pos; i++) {
    if (bad[pos][i]) continue;
    arr[pos] = i;
    rec(pos + 1);
    arr[pos] = 0;
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    in[y].push_back(i);
    out[x].push_back({z, i});
  }
  for (int i = 1; i <= n; i++) {
    sort(begin(out[i]), end(out[i]));
    for (int j = 1; j <= (int)(out[i]).size(); j++) {
      int ind = out[i][j - 1][1];
      pt[ind] = {(int)(out[i]).size(), j};
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int ind : in[i]) {
      int x = pt[ind][0], y = pt[ind][1];
      cnt[x][y]++;
    }
    for (int ind : in[i]) {
      int x = pt[ind][0], y = pt[ind][1];
      if (cnt[x][y] == 1) {
        groups[x][y][i] = 1;
      } else {
        bad[x][y] = 1;
      }
    }
    for (int ind : in[i]) {
      int x = pt[ind][0], y = pt[ind][1];
      cnt[x][y]--;
    }
  }
  for (int x = 1; x <= k; x++)
    for (int y = 1; y <= x; y++)
      for (int x2 = x + 1; x2 <= k; x2++)
        for (int y2 = 1; y2 <= x2; y2++)
          if ((groups[x][y] & groups[x2][y2]).count()) {
            badpair[x][y][x2][y2] = 1;
          }
  rec(1);
  cout << ans;
}
