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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const int C = 2e7 + 10, N = 2e5 + 10;
int mod;
const int K = 9;
int p[C];
int h[K][K];
int need, ans;
int n, m, k;
vector<int> primes;
vector<pair<int, int> > g[N];
void calc() {
  auto is_prime = [&](int x) {
    for (int i = 2; i * i <= x; i++)
      if (x % i == 0) return false;
    return true;
  };
  mod = rnd() % (int)(1e9 + 7) + (int)1e8;
  while (!is_prime(mod)) mod++;
  for (int i = 2; i < C; i++) {
    if (p[i]) continue;
    primes.push_back(i);
    for (int j = i * 2; j < C; j += i) p[j] = 1;
  }
}
void rec(int pos, int msk) {
  if (pos == k) {
    if (msk == need) ans++;
    return;
  }
  for (int i = 0; i <= pos; i++) rec(pos + 1, 1ll * msk * h[pos][i] % mod);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  calc();
  cin >> n >> m >> k;
  while (m--) {
    int x, y, w;
    cin >> x >> y >> w;
    g[--x].push_back({--y, w});
  }
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++) h[i][j] = 1;
  for (int v = 0; v < n; v++) {
    sort((g[v]).begin(), (g[v]).end(), [&](pair<int, int> a, pair<int, int> b) {
      return a.second < b.second;
    });
    int pw = (int)(g[v]).size() - 1;
    for (int i = 0; i < (int)(g[v]).size(); i++) {
      int to = g[v][i].first;
      h[pw][i] = 1ll * h[pw][i] * primes[to] % mod;
    }
  }
  need = 1;
  for (int i = 0; i < n; i++) need = 1ll * need * primes[i] % mod;
  rec(0, 1);
  cout << ans;
  return 0;
}
