#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
const int K = 10;
int cnt[N], w[10], deg[N], nGd[K][K][K][K], chk[K][K];
vector<int> G[K];
vector<pair<int, int> > E[N];
bitset<N> U[K][K];
int n, m, k, res;
void dfs(int x) {
  if (x == k + 1) {
    ++res;
    return;
  }
  for (int i = 1; i <= x; ++i) {
    if (chk[x][i]) continue;
    int ok = 0;
    for (int j = 1; j < x; ++j) {
      if (nGd[x][i][j][w[j]]) {
        ok = 1;
        break;
      }
    }
    if (ok) continue;
    w[x] = i;
    dfs(x + 1);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; ++i) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    E[x].push_back(make_pair(z, y));
    ++deg[x];
  }
  for (int i = 1; i <= n; ++i) {
    sort(E[i].begin(), E[i].end());
    G[deg[i]].push_back(i);
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= i; ++j) {
      U[i][j].reset();
      for (int ii = 1; ii <= n; ++ii) cnt[ii] = 0;
      for (auto& x : G[i]) {
        int y = E[x][j - 1].second;
        ++cnt[y];
        if (cnt[y] > 1) chk[i][j] = 1;
        U[i][j][y] = 1;
      }
    }
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= i; ++j) {
      for (int ii = i + 1; ii <= k; ++ii) {
        for (int jj = 1; jj <= ii; ++jj) {
          if ((U[i][j] & U[ii][jj]).count()) {
            nGd[i][j][ii][jj] = nGd[ii][jj][i][j] = 1;
          }
        }
      }
    }
  }
  dfs(1);
  printf("%d", res);
  return 0;
}

### 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;
const int N = 200005;
const int K = 10;
int cnt[N], w[10], deg[N], nGd[K][K][K][K], chk[K][K];
vector<int> G[K];
vector<pair<int, int> > E[N];
bitset<N> U[K][K];
int n, m, k, res;
void dfs(int x) {
  if (x == k + 1) {
    ++res;
    return;
  }
  for (int i = 1; i <= x; ++i) {
    if (chk[x][i]) continue;
    int ok = 0;
    for (int j = 1; j < x; ++j) {
      if (nGd[x][i][j][w[j]]) {
        ok = 1;
        break;
      }
    }
    if (ok) continue;
    w[x] = i;
    dfs(x + 1);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; ++i) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    E[x].push_back(make_pair(z, y));
    ++deg[x];
  }
  for (int i = 1; i <= n; ++i) {
    sort(E[i].begin(), E[i].end());
    G[deg[i]].push_back(i);
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= i; ++j) {
      U[i][j].reset();
      for (int ii = 1; ii <= n; ++ii) cnt[ii] = 0;
      for (auto& x : G[i]) {
        int y = E[x][j - 1].second;
        ++cnt[y];
        if (cnt[y] > 1) chk[i][j] = 1;
        U[i][j][y] = 1;
      }
    }
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= i; ++j) {
      for (int ii = i + 1; ii <= k; ++ii) {
        for (int jj = 1; jj <= ii; ++jj) {
          if ((U[i][j] & U[ii][jj]).count()) {
            nGd[i][j][ii][jj] = nGd[ii][jj][i][j] = 1;
          }
        }
      }
    }
  }
  dfs(1);
  printf("%d", res);
  return 0;
}

```