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 maxn = 2e5 + 10;
const int inf = 1e9;
vector<pair<int, int>> g[maxn], deg[maxn];
bool cant[10][10][10][10];
int p[10];
long long answer = 0;
void bt(int now, int k) {
  if (now == k + 1) {
    bool good = 1;
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        if (cant[i][p[i]][j][p[j]]) good = 0;
    answer += good;
    return;
  }
  for (int i = 1; i <= now; i++) {
    p[now] = i;
    bt(now + 1, k);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    g[v].push_back({w, u});
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    int k = g[i].size();
    for (int j = 0; j < k; j++) {
      int u = g[i][j].second;
      deg[u].push_back({k, j + 1});
    }
  }
  for (int u = 1; u <= n; u++) {
    sort(deg[u].begin(), deg[u].end());
    for (int j = 1; j < deg[u].size(); j++) {
      if (deg[u][j] == deg[u][j - 1]) {
        int k = deg[u][j].first, x = deg[u][j].second;
        cant[k][x][k][x] = 1;
      }
    }
    deg[u].resize(unique(deg[u].begin(), deg[u].end()) - deg[u].begin());
    for (int j = 1; j < deg[u].size(); j++) {
      for (int z = 0; z < j; z++) {
        int k = deg[u][j].first, x = deg[u][j].second;
        int K = deg[u][z].first, X = deg[u][z].second;
        cant[k][x][K][X] = 1;
      }
    }
  }
  bt(1, k);
  cout << answer << endl;
}
