#include <bits/stdc++.h>
using namespace std;
int n, m, k;
struct par {
  int y, z;
  par(int yy = 0, int zz = 0) : y(yy), z(zz) {}
  bool operator<(const par &B) const { return z < B.z; }
};
vector<par> e[200010];
int val[200010], S[10][10], ans = 0, aim = 1;
void dfs(int x, int now) {
  if (x > k) {
    if (now == aim) ans++;
    return;
  }
  for (int i = 1; i <= x; i++) dfs(x + 1, 1ll * now * S[x][i] % 1000000007);
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1, x, y, z; i <= m; i++) {
    scanf("%d %d %d", &x, &y, &z);
    e[x].push_back(par(y, z));
  }
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) S[i][j] = 1;
  for (int i = 1; i <= n; i++)
    val[i] = ((i << 3) + (i >> 2) * 173 + 23333333) % 99049904,
    aim = 1ll * aim * val[i] % 1000000007;
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end());
    for (int j = 0; j < e[i].size(); j++)
      S[e[i].size()][j + 1] =
          1ll * S[e[i].size()][j + 1] * val[e[i][j].y] % 1000000007;
  }
  dfs(1, 1);
  printf("%d", ans);
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, k;
struct par {
  int y, z;
  par(int yy = 0, int zz = 0) : y(yy), z(zz) {}
  bool operator<(const par &B) const { return z < B.z; }
};
vector<par> e[200010];
int val[200010], S[10][10], ans = 0, aim = 1;
void dfs(int x, int now) {
  if (x > k) {
    if (now == aim) ans++;
    return;
  }
  for (int i = 1; i <= x; i++) dfs(x + 1, 1ll * now * S[x][i] % 1000000007);
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1, x, y, z; i <= m; i++) {
    scanf("%d %d %d", &x, &y, &z);
    e[x].push_back(par(y, z));
  }
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) S[i][j] = 1;
  for (int i = 1; i <= n; i++)
    val[i] = ((i << 3) + (i >> 2) * 173 + 23333333) % 99049904,
    aim = 1ll * aim * val[i] % 1000000007;
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end());
    for (int j = 0; j < e[i].size(); j++)
      S[e[i].size()][j + 1] =
          1ll * S[e[i].size()][j + 1] * val[e[i][j].y] % 1000000007;
  }
  dfs(1, 1);
  printf("%d", ans);
}

```