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 INF = 1000000007;
const long long MOD = 1000000007;
const int N = 200005;
const int K = 10;
int n, m, k;
vector<pair<int, int> > graph[N];
int arr[N][K][K];
int in[K][K];
vector<pair<int, int> > g[K][K];
int ans = 0;
void gen(int x) {
  if (x == k) {
    ans++;
    return;
  }
  for (int i = 0; i <= x; i++) {
    if (in[x][i]) continue;
    for (pair<int, int> p : g[x][i]) {
      in[p.first][p.second]++;
    }
    gen(x + 1);
    for (pair<int, int> p : g[x][i]) {
      in[p.first][p.second]--;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    graph[u].push_back({w, v});
  }
  for (int i = 0; i < n; i++) {
    sort(graph[i].begin(), graph[i].end());
    int l = graph[i].size();
    for (int j = 0; j < l; j++) {
      arr[graph[i][j].second][l - 1][j]++;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j1 = 0; j1 < k; j1++) {
      for (int j2 = 0; j2 <= j1; j2++) {
        in[j1][j2] += arr[i][j1][j2] >= 2;
        for (int j3 = 0; j3 < j1; j3++) {
          for (int j4 = 0; j4 <= j3; j4++) {
            if (arr[i][j3][j4] && arr[i][j1][j2]) {
              g[j3][j4].push_back({j1, j2});
            }
          }
        }
      }
    }
  }
  gen(0);
  cout << ans;
  return 0;
}
