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 MX = 300005;
const long long MOD = 1000000007;
void setIO(string name) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
int t;
int n, k, m;
bool works[10][10][10][10];
pair<int, int> wt[MX];
vector<int> nums[10][10];
int deg[MX], deg1[MX];
bool v[MX];
int perm[10], fact[10];
bool check(int a, int x, int b, int y) {
  for (int i = 0; i < (m + 1); i++) v[i] = 0;
  for (int i = 0; i < (nums[a][x].size()); i++) {
    if (v[nums[a][x][i]] == 1) return 0;
    v[nums[a][x][i]] = 1;
  }
  if (a == b) return 1;
  for (int i = 0; i < (nums[b][y].size()); i++) {
    if (v[nums[b][y][i]] == 1) return 0;
    v[nums[b][y][i]] = 1;
  }
  return 1;
}
void solve() {
  cin >> n >> m >> k;
  for (int i = (1); i < (n + 1); i++) {
    deg[i] = 0;
    deg1[i] = 0;
  }
  for (int i = 0; i < (m); i++) {
    int u, v, w;
    cin >> u >> v >> w;
    wt[w] = make_pair(u, v);
    deg[u]++;
  }
  for (int i = (1); i < (m + 1); i++) {
    pair<int, int> p = wt[i];
    deg1[p.first]++;
    nums[deg[p.first]][deg1[p.first]].push_back(p.second);
  }
  for (int i = (1); i < (k + 1); i++) {
    for (int i1 = (1); i1 < (i + 1); i1++) {
      for (int j = (1); j < (k + 1); j++) {
        for (int j1 = (1); j1 < (j + 1); j1++) {
          works[i][i1][j][j1] = check(i, i1, j, j1);
        }
      }
    }
  }
  fact[0] = 1;
  for (int i = (1); i < (k + 1); i++) fact[i] = (i * fact[i - 1]);
  int tot = 0;
  for (int val = 0; val < (fact[k]); val++) {
    int temp = val;
    for (int i = (k + 1) - 1; i >= (1); i--) {
      perm[i] = temp / fact[i - 1] + 1;
      temp = (temp % fact[i - 1]);
    }
    bool gah = 1;
    for (int i = (1); i < (k + 1); i++) {
      for (int j = (i); j < (k + 1); j++) {
        gah = (gah && works[i][perm[i]][j][perm[j]]);
      }
    }
    if (gah) {
      tot++;
    }
  }
  cout << tot << endl;
}
int main() {
  t = 1;
  for (int i = 0; i < (t); i++) {
    solve();
  }
}
