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 INF = 2147483647;
const long long LLINF = 9223372036854775807LL;
const int maxn = 200010;
vector<pair<int, int> > a[maxn];
const int mod = 1000000009;
int hh[maxn];
int get_hash(int x) { return hh[x]; }
int th[10][10];
int target_hash = 0;
int result = 0;
int cvals[10];
void rec(int pos, int k, int cur_sum) {
  if (pos == k) {
    if (cur_sum == target_hash) result += 1;
    return;
  }
  for (int i = 0; i < pos + 1; ++i) {
    cvals[pos] = i;
    int new_sum = (cur_sum + th[pos + 1][i]) % mod;
    rec(pos + 1, k, new_sum);
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n + 1; ++i) {
    hh[i] = ((long long)i * i + rand()) % mod;
  }
  for (int i = 0; i < m; ++i) {
    int fr, to, w;
    scanf("%d%d%d", &fr, &to, &w);
    --fr, --to;
    a[fr].push_back(make_pair(w, to));
  }
  for (int i = 0; i < n; ++i) sort(a[i].begin(), a[i].end());
  for (int i = 1; i <= n; ++i) target_hash = (target_hash + get_hash(i)) % mod;
  for (int i = 0; i < n; ++i) {
    int cur_deg = int((a[i]).size());
    for (int j = 0; j < cur_deg; ++j) {
      int cur_h = get_hash(a[i][j].second + 1);
      th[cur_deg][j] = (th[cur_deg][j] + cur_h) % mod;
    }
  }
  rec(0, k, 0);
  cout << result << endl;
  return 0;
}
