#include <bits/stdc++.h>
using namespace std;
int min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
void bmax(int &a, int b) { a = (a > b) ? a : b; }
void bmin(int &a, int b) { a = (a < b) ? a : b; }
const int L = 1 << 20;
char buf[L], *p1, *p2;
int read() {
  int x = 0, w = 1;
  char ch = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, L, stdin),
                         p1 == p2 ? EOF : *p1++)
                      : *p1++);
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, L, stdin),
                      p1 == p2 ? EOF : *p1++)
                   : *p1++);
  }
  while (ch <= '9' && ch >= '0')
    x = x * 10 + ch - '0',
    ch = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, L, stdin),
                      p1 == p2 ? EOF : *p1++)
                   : *p1++);
  return x * w;
}
const int N = 2e5 + 5, base = 13331;
vector<pair<int, int> > g[N];
unsigned long long las, a[20][20], powe[N];
int n, m, k, ans;
void DFS(int u, unsigned long long nw) {
  if (u == k + 1) {
    ans += (nw == las);
    return;
  }
  for (int i = 1; i <= u; i++) DFS(u + 1, nw + a[u][i]);
}
int main() {
  n = read();
  m = read();
  k = read();
  for (int i = 1, u, v, w; i <= m; i++)
    u = read(), v = read(), w = read(), g[u].push_back(make_pair(w, v));
  powe[0] = 1;
  las = 0;
  for (int i = 1; i <= n; i++) powe[i] = powe[i - 1] * base, las += powe[i];
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    int tot = g[i].size();
    for (int j = 0; j < tot; j++) a[tot][j + 1] += powe[g[i][j].second];
  }
  ans = 0;
  DFS(1, 0);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 min(int a, int b) { return a < b ? a : b; }
int max(int a, int b) { return a > b ? a : b; }
void bmax(int &a, int b) { a = (a > b) ? a : b; }
void bmin(int &a, int b) { a = (a < b) ? a : b; }
const int L = 1 << 20;
char buf[L], *p1, *p2;
int read() {
  int x = 0, w = 1;
  char ch = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, L, stdin),
                         p1 == p2 ? EOF : *p1++)
                      : *p1++);
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, L, stdin),
                      p1 == p2 ? EOF : *p1++)
                   : *p1++);
  }
  while (ch <= '9' && ch >= '0')
    x = x * 10 + ch - '0',
    ch = (p1 == p2 ? (p2 = buf + fread(p1 = buf, 1, L, stdin),
                      p1 == p2 ? EOF : *p1++)
                   : *p1++);
  return x * w;
}
const int N = 2e5 + 5, base = 13331;
vector<pair<int, int> > g[N];
unsigned long long las, a[20][20], powe[N];
int n, m, k, ans;
void DFS(int u, unsigned long long nw) {
  if (u == k + 1) {
    ans += (nw == las);
    return;
  }
  for (int i = 1; i <= u; i++) DFS(u + 1, nw + a[u][i]);
}
int main() {
  n = read();
  m = read();
  k = read();
  for (int i = 1, u, v, w; i <= m; i++)
    u = read(), v = read(), w = read(), g[u].push_back(make_pair(w, v));
  powe[0] = 1;
  las = 0;
  for (int i = 1; i <= n; i++) powe[i] = powe[i - 1] * base, las += powe[i];
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    int tot = g[i].size();
    for (int j = 0; j < tot; j++) a[tot][j + 1] += powe[g[i][j].second];
  }
  ans = 0;
  DFS(1, 0);
  printf("%d\n", ans);
  return 0;
}

```