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>
const unsigned long long M1 = 1e8 + 7;
const unsigned long long M2 = 1e9 + 7;
const unsigned long long M3 = 998244353;
const int N = 2e5 + 5;
int n, m, k, U[N], V[N], h[10], ans;
unsigned long long k1[N], k2[N], k3[N], f1[10][10], f2[10][10], f3[10][10],
    g[10][10];
unsigned long long s1, s2, s3;
std::vector<int> E[N];
void dfs(int dep) {
  if (dep == k + 1) {
    unsigned long long b1, b2, b3, d = 0;
    b1 = 0;
    b2 = 0;
    b3 = 0;
    for (int i = 1; i <= k; i++) {
      b1 += f1[i][h[i]];
      b2 += f2[i][h[i]];
      b3 += f3[i][h[i]];
      d += g[i][h[i]];
    }
    if (b1 == s1 && b2 == s2 && b3 == s3 && d == n) ++ans;
    return;
  }
  for (int i = 1; i <= dep; i++) {
    h[dep] = i;
    dfs(dep + 1);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1, x, y, w; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &w);
    U[w] = x;
    V[w] = y;
    E[x].push_back(w);
  }
  k1[0] = k2[0] = k3[0] = 1;
  for (int i = 1; i <= n; i++) {
    std::sort(E[i].begin(), E[i].end());
    k1[i] = k1[i - 1] * M1;
    k2[i] = k2[i - 1] * M2;
    k3[i] = k3[i - 1] * M3;
    s1 += k1[i];
    s2 += k2[i];
    s3 += k3[i];
  }
  for (int i = 1; i <= n; i++)
    for (int v = 0, s = E[i].size(); v < s; v++) {
      f1[s][v + 1] += k1[V[E[i][v]]];
      f2[s][v + 1] += k2[V[E[i][v]]];
      f3[s][v + 1] += k3[V[E[i][v]]];
      g[s][v + 1]++;
    }
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
