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;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const long long INF = 1e18;
const long long MAX_N = 200000;
vector<pair<long long, long long> > edges[MAX_N];
vector<long long> inv[MAX_N];
bitset<MAX_N> state[9][9];
bool BANNED[9][9][9][9];
bool NG[9][9];
vector<long long> v;
long long N, M, K;
long long ans;
void f() {
  for (int i = 0; i < K; i++) {
    if (NG[i][v[i]]) return;
    for (int j = i + 1; j < K; j++) {
      if (BANNED[i][v[i]][j][v[j]]) return;
    }
  }
  ans++;
}
void dfs(int idx) {
  if (idx == K) {
    f();
    return;
  }
  for (int i = 0; i <= idx; i++) {
    v[idx] = i;
    dfs(idx + 1);
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N >> M >> K;
  v.resize(K);
  for (int i = 0; i < M; i++) {
    long long a, b, w;
    cin >> a >> b >> w;
    a--;
    b--;
    edges[a].push_back({w, b});
    inv[b].push_back(a);
  }
  for (int i = 0; i < N; i++) {
    sort(edges[i].begin(), edges[i].end());
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < edges[i].size(); j++) {
      int to = edges[i][j].second;
      if (state[edges[i].size() - 1][j][to]) NG[edges[i].size() - 1][j] = true;
      state[edges[i].size() - 1][j][to] = 1;
    }
  }
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= i; j++) {
      for (int k = i; k < K; k++) {
        for (int l = 0; l <= k; l++) {
          bitset<MAX_N> a = state[i][j] & state[k][l];
          if (a.count() >= 1) {
            BANNED[i][j][k][l] = true;
          }
        }
      }
    }
  }
  dfs(0);
  cout << ans << "\n";
  return 0;
}
