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;
inline int rd() {
  int x = 0;
  bool f = 0;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}
struct edge {
  int to, w;
} e[200007][9];
inline bool cmp(edge a, edge b) { return a.w < b.w; }
int n, m, k, ans, deg[200007];
long long tot, pw[200007], sum[20][20];
long long tot2, pw2[200007], sum2[20][20];
void dfs(int t, long long s, long long s2) {
  if (t == k + 1) {
    if (s == tot && s2 == tot2) ++ans;
    return;
  }
  for (int i = 1; i <= t; ++i) dfs(t + 1, s + sum[t][i], s2 + sum2[t][i]);
}
int main() {
  n = rd();
  m = rd();
  k = rd();
  for (int i = 1, u, v, w; i <= m; ++i) {
    u = rd();
    v = rd();
    w = rd();
    e[u][++deg[u]].to = v;
    e[u][deg[u]].w = w;
  }
  pw[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw[i] = 1ll * pw[i - 1] * n % 998244353ll;
    tot += pw[i];
  }
  pw2[0] = 1;
  for (int i = 1; i <= n; ++i) {
    pw2[i] = 1ll * pw2[i - 1] * n % 19260817ll;
    tot2 += pw2[i];
  }
  for (int i = 1; i <= n; ++i) {
    sort(e[i] + 1, e[i] + 1 + deg[i], cmp);
    for (int j = 1; j <= deg[i]; ++j) {
      sum[deg[i]][j] += pw[e[i][j].to];
      sum2[deg[i]][j] += pw2[e[i][j].to];
    }
  }
  dfs(1, 0, 0);
  printf("%d\n", ans);
  return 0;
}
