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 MAXN = 1e6 + 5;
int visited[10][10];
int n, m, K;
vector<pair<int, int>> blockat[10][10];
vector<pair<int, int>> v1[MAXN];
bool blocked[MAXN][10][10];
int solve(int x) {
  if (x == (K + 1)) {
    return 1;
  }
  int ans = 0;
  for (int i = 1; i <= x; i++) {
    if (visited[x][i]) {
      continue;
    }
    for (auto y : blockat[x][i]) {
      visited[y.first][y.second]++;
    }
    ans += solve(x + 1);
    for (auto y : blockat[x][i]) {
      visited[y.first][y.second]--;
    }
  }
  return ans;
}
int main() {
  cin >> n >> m >> K;
  for (int i = 1; i <= m; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    v1[x].push_back(make_pair(w, y));
  }
  for (int i = 1; i <= n; i++) {
    sort(v1[i].begin(), v1[i].end());
    for (int j = 0; j < v1[i].size(); j++) {
      if (blocked[v1[i][j].second][(int)v1[i].size()][j + 1]) {
        visited[(int)v1[i].size()][j + 1]++;
      }
      blocked[v1[i][j].second][(int)v1[i].size()][j + 1] = true;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= K; j++) {
      for (int k = 1; k < j; k++) {
        for (int x = 1; x <= j; x++) {
          for (int y = 1; y <= k; y++) {
            if (blocked[i][j][x] && blocked[i][k][y]) {
              blockat[k][y].push_back(make_pair(j, x));
            }
          }
        }
      }
    }
  }
  cout << solve(1) << endl;
}
