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;
int n, m, k, d, i, f[10][10][10][10], a[105], ans, u, v, w, j, l, A, y[15][15];
struct str {
  int v, w;
};
vector<str> g[200005];
bitset<200005> p[10][10];
bool cmp(str a, str b) { return a.w < b.w; }
void dfs(int i) {
  if (i > A) {
    int j, k;
    for (j = 1; j <= A; ++j)
      for (k = j + 1; k <= A; ++k)
        if (f[j][a[j]][k][a[k]]) return;
    ++ans;
    return;
  }
  int j;
  for (j = 0; j < i; ++j)
    if (!y[i][j]) {
      a[i] = j;
      dfs(i + 1);
    }
}
int main() {
  scanf("%d %d %d", &n, &m, &A);
  for (i = 1; i <= m; ++i) {
    scanf("%d %d %d", &u, &v, &w);
    g[u].push_back((str){v, w});
  }
  for (i = 1; i <= n; ++i) {
    sort(g[i].begin(), g[i].end(), cmp);
    for (j = 0; j < g[i].size(); ++j) {
      if (p[g[i].size()][j][g[i][j].v]) y[g[i].size()][j] = 1;
      p[g[i].size()][j][g[i][j].v] = 1;
    }
  }
  for (i = 1; i <= A; ++i)
    for (j = 0; j < i; ++j)
      for (k = i + 1; k <= A; ++k)
        for (l = 0; l < k; ++l)
          if ((p[i][j] & p[k][l]).any()) f[i][j][k][l] = 1;
  dfs(1);
  cout << ans;
}
