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;
int n, m, k;
vector<vector<pair<int, int>>> g;
vector<vector<pair<int, int>>> ag;
int can[9][9];
int zap[9][9][9][9];
void count_all() {
  for (int i = 0; i < n; i++) {
    int my[9][9];
    for (int j = 0; j < 9; j++) {
      for (int y = 0; y < 9; y++) {
        my[j][y] = 0;
      }
    }
    for (auto p : ag[i]) {
      int w = p.first;
      int u = p.second;
      int k1 = g[u].size();
      int p1 =
          upper_bound((g[u]).begin(), (g[u]).end(), (pair<int, int>){w, -1}) -
          g[u].begin();
      my[k1 - 1][p1]++;
    }
    vector<pair<int, int>> m1;
    for (int j = 0; j < k; j++) {
      for (int y = 0; y < k; y++) {
        if (my[j][y] > 1) {
          can[j][y] = 1;
        }
        if (my[j][y] == 1) {
          m1.push_back({j, y});
        }
      }
    }
    for (int j = 0; j < m1.size(); j++) {
      for (int y = j + 1; y < m1.size(); y++) {
        zap[m1[j].first][m1[j].second][m1[y].first][m1[y].second] = 1;
      }
    }
  }
}
long long ans;
void solve(int c) {
  if (c == k) {
    ans++;
    return;
  }
  for (int i = 0; i <= c; i++) {
    if (can[c][i] == 0) {
      for (int j = c + 1; j < k; j++) {
        for (int f = 0; f <= j; f++) {
          can[j][f] += zap[c][i][j][f];
        }
      }
      solve(c + 1);
      for (int j = c + 1; j < k; j++) {
        for (int f = 0; f <= j; f++) {
          can[j][f] -= zap[c][i][j][f];
        }
      }
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  g.resize(n);
  ag.resize(n);
  for (int i = 0; i < m; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    a--, b--;
    g[a].push_back({w, b});
    ag[b].push_back({w, a});
  }
  count_all();
  solve(0);
  cout << ans << '\n';
  return 0;
}
