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 N = 2e5 + 100;
long long id[10][10];
vector<pair<int, int>> G[N];
vector<int> rG[N];
int on[105];
long long B[105];
int ans;
int n, m, K;
void solve(int pos, long long mask, int cnt) {
  if (pos > K) {
    if (cnt == n) ans++;
    return;
  }
  for (int i = 1; i <= pos; i++) {
    if ((mask >> id[pos][i]) & 1LL) continue;
    solve(pos + 1, mask | B[id[pos][i]], cnt + on[id[pos][i]]);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> K;
  int idx = 0;
  for (int i = 1; i <= K; i++) {
    for (int j = 1; j <= i; j++) {
      id[i][j] = idx++;
    }
  }
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    G[u].push_back({w, v});
    rG[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    sort(G[i].begin(), G[i].end());
  }
  vector<vector<int>> cnt(K + 1, vector<int>(K + 1));
  long long blocked = 0;
  for (int i = 1; i <= n; i++) {
    for (int u : rG[i]) {
      for (int j = 0; j < G[u].size(); j++) {
        if (G[u][j].second == i) {
          cnt[G[u].size()][j + 1]++;
          break;
        }
      }
    }
    long long mask = 0;
    for (int k = 1; k <= K; k++) {
      for (int l = 1; l <= k; l++) {
        if (cnt[k][l] > 1) {
          blocked |= (1LL << id[k][l]);
        } else if (cnt[k][l]) {
          mask |= (1LL << id[k][l]);
          on[id[k][l]]++;
        }
      }
    }
    for (long long k = 0; k < idx; k++) {
      if ((mask >> k) & 1LL) {
        B[k] |= mask;
      }
    }
    for (int k = 1; k <= K; k++) {
      for (int l = 1; l <= k; l++) {
        cnt[k][l] = 0;
      }
    }
  }
  solve(1, blocked, 0);
  cout << ans << endl;
  return 0;
}
