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>
#pragma GCC optimize("Ofast")
using namespace std;
mt19937 mt_rand(time(0));
vector<pair<long long, long long>> g[200005];
long long num[200005], ha[25][25] = {0}, sum = 0, ans = 0, n, m, k;
void DFS(long long now, long long v) {
  if (now == k) {
    ans += (v == sum);
    return;
  }
  int i;
  for (i = 0; i <= now; i++) DFS(now + 1, v ^ ha[now][i]);
}
int main() {
  long long i, j, from, to, cost;
  scanf("%lld%lld%lld", &n, &m, &k);
  for (i = 0; i < m; i++) {
    scanf("%lld%lld%lld", &from, &to, &cost);
    g[from - 1].emplace_back(cost, to - 1);
  }
  for (i = 0; i < n; i++) {
    num[i] = mt_rand();
    sum ^= num[i];
  }
  for (i = 0; i < n; i++) {
    sort(g[i].begin(), g[i].end());
    for (j = 0; j < g[i].size(); j++)
      ha[g[i].size() - 1][j] ^= num[g[i][j].second];
  }
  DFS(0, 0);
  printf("%lld\n", ans);
}
