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 + 5, HS = 3, K = 10;
int n, m, k;
vector<pair<int, int> > g[N];
int mod[HS];
struct hash_number {
  int a[HS];
  hash_number() { fill(a, a + HS, 0); }
  hash_number(long long x) {
    for (int i = (0); i <= (HS - 1); ++i) a[i] = (x % mod[i] + mod[i]) % mod[i];
  }
  hash_number operator+(hash_number x) {
    hash_number res;
    res.a[0] = (a[0] + x.a[0]) % mod[0];
    res.a[1] = (a[1] * 1ll * x.a[1]) % mod[1];
    res.a[2] = (a[2] + x.a[2]) % mod[2];
    return res;
  }
  bool operator==(const hash_number& x) const {
    for (int i = (0); i <= (HS - 1); ++i)
      if (a[i] != x.a[i]) return 0;
    return 1;
  }
} val[N], c[K][K], s;
int ans;
int status[K];
void dfs(int x, hash_number hsh) {
  if (x == k) {
    if (hsh == s) ++ans;
    return;
  }
  for (int i = (1); i <= (x + 1); ++i) {
    status[x + 1] = i;
    dfs(x + 1, hsh + c[x + 1][i]);
  }
}
int main() {
  mod[0] = 998244353;
  mod[1] = 1e9 + 7;
  mod[2] = 2e8 + 11;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = (1); i <= (m); ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    g[u].push_back({w, v});
  }
  for (int i = (1); i <= (n); ++i) val[i] = hash_number(rand());
  for (int i = (1); i <= (n); ++i) s = s + val[i];
  for (int u = (1); u <= (n); ++u) {
    int d = g[u].size();
    sort(g[u].begin(), g[u].end());
    for (int i = 1; i <= g[u].size(); ++i) {
      int v = g[u][i - 1].second;
      c[d][i] = c[d][i] + val[v];
    }
  }
  dfs(0, hash_number());
  printf("%d\n", ans);
  return 0;
}
