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>
#pragma GCC optimize("O2")
#pragma GCC target("avx2")
using namespace std;
const char nl = '\n';
const long long inf = 1e7;
const long long inf2 = 1e18 + 10;
const long double inf3 = 1e14;
const long long mod = 1e9 + 7;
const long double eps = 1e-6;
const int logn = 19;
const int maxn = 200000, maxm = 1e6;
vector<pair<int, int>> g[maxn];
int n, m, k, ans;
vector<int> a[10];
int p[10];
bool gd[10][10][10][10], cnt[maxn];
void solve(int x) {
  if (x > k) {
    ans++;
    return;
  }
  for (int j = 1; j <= x; j++) {
    bool good = true;
    p[x] = j;
    for (int i = 1; i <= x; i++) {
      good = good && gd[i][p[i]][x][j];
    }
    if (good) {
      solve(x + 1);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u - 1].emplace_back(w, v - 1);
  }
  for (int i = 0; i < n; i++) {
    sort((g[i]).begin(), (g[i]).end());
  }
  for (int i = 0; i < n; i++) {
    a[(int)g[i].size()].push_back(i);
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int x = i + 1; x <= k; x++) {
        for (int y = 1; y <= x; y++) {
          for (int v : a[i]) {
            cnt[g[v][j - 1].second] = true;
          }
          bool good = true;
          for (int v : a[x]) {
            if (cnt[g[v][y - 1].second]) {
              good = false;
              break;
            }
            cnt[g[v][y - 1].second] = true;
          }
          gd[i][j][x][y] = good;
          gd[x][y][i][j] = good;
          for (int v : a[i]) {
            cnt[g[v][j - 1].second] = false;
          }
          for (int v : a[x]) {
            cnt[g[v][y - 1].second] = false;
          }
        }
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      bool good = true;
      for (int v : a[i]) {
        if (cnt[g[v][j - 1].second]) {
          good = false;
          break;
        }
        cnt[g[v][j - 1].second] = true;
      }
      gd[i][j][i][j] = good;
      for (int v : a[i]) {
        cnt[g[v][j - 1].second] = false;
      }
    }
  }
  solve(1);
  cout << ans;
}
