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 N = 200000;
const int HashMOD = 233333;
const int HashBASE = 13;
vector<int> jh[10];
struct Node {
  int to, w;
};
vector<Node> edge[N + 50];
int n, m, out[N + 50], hashzhi[N + 50], ans, anss, k, sum[10][10];
void Read(int &x) {
  x = 0;
  int p = 0;
  char st = getchar();
  while (st < '0' || st > '9') p = (st == '-'), st = getchar();
  while (st >= '0' && st <= '9')
    x = (x << 1) + (x << 3) + st - '0', st = getchar();
  x = p ? -x : x;
  return;
}
int Cmp(Node a, Node b) { return a.w < b.w; }
int Solve(int x, int y) {
  int ans = 0;
  for (vector<int>::iterator it = jh[x].begin(); it != jh[x].end(); it++) {
    int v = *it;
    ans = (0LL + ans + hashzhi[edge[v][y - 1].to]) % HashMOD;
  }
  return ans;
}
void Dfs(int t, int tmp) {
  if (t == k + 1) {
    if (tmp == anss) ans++;
    return;
  }
  for (int i = 1; i <= t; i++) Dfs(t + 1, (tmp + sum[t][i]) % HashMOD);
  return;
}
int main() {
  Read(n);
  Read(m);
  Read(k);
  hashzhi[0] = 1;
  for (int i = 1; i <= n; i++)
    hashzhi[i] = 1LL * hashzhi[i - 1] * HashBASE % HashMOD,
    anss = (0LL + hashzhi[i] + anss) % HashMOD;
  for (int i = 1, u, v, w; i <= m; i++) {
    Read(u);
    Read(v);
    Read(w);
    edge[u].push_back((Node){v, w});
    out[u]++;
  }
  for (int i = 1; i <= n; i++)
    jh[out[i]].push_back(i), sort(edge[i].begin(), edge[i].end(), Cmp);
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) sum[i][j] = Solve(i, j);
  Dfs(1, 0);
  printf("%d", ans);
  return 0;
}
