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;
long long power(long long a, long long b) {
  long long res = 1;
  a = a % 1000000007;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 1000000007;
      b--;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 1000000007 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
vector<pair<int, int>> adj[200001];
vector<int> d(200001, 0);
vector<int> d1[10];
int n, k;
bitset<200001> b[82];
vector<int> c1(82);
int dp[82][82];
long long visit(int ctr1[], int ctr, int ctr2) {
  if (ctr == 0 && ctr2 == n) return 1;
  int pos = (ctr - 1) * 9;
  long long ans = 0ll;
  for (int i = 0; i < ctr; i++) {
    int check = 1;
    for (int j = ctr; j < k; j++) {
      if (dp[(ctr - 1) * 9 + i][j * 9 + ctr1[j]] == 0) check = 0;
    }
    if (check == 1) {
      ctr1[ctr - 1] = i;
      int ctr3 = ctr2 + c1[(ctr - 1) * 9 + i];
      ans += visit(ctr1, ctr - 1, ctr3);
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    int m;
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++) {
      int u, v, w;
      cin >> u >> v >> w;
      adj[u].push_back(make_pair(w, v));
      d[u]++;
    }
    for (int i = 1; i <= n; i++) {
      sort(adj[i].begin(), adj[i].end());
      d1[d[i]].push_back(i);
    }
    for (int i = 1; i <= k; i++) {
      int pos = (i - 1) * 9;
      for (int j = 0; j < i; j++) {
        pos += j;
        int f = d1[i].size();
        for (int l = 0; l < f; l++) b[pos][adj[d1[i][l]][j].second] = 1;
        pos -= j;
      }
    }
    for (int i = 0; i <= 81; i++)
      for (int j = 0; j <= 81; j++) dp[i][j] = 0;
    for (int i = 0; i < 82; i++) c1[i] = b[i].count();
    for (int i = 0; i < k; i++) {
      for (int j = 0; j <= i; j++) {
        for (int l = i + 1; l < k; l++) {
          for (int h = 0; h <= l; h++) {
            b[i * 9 + j] ^= b[l * 9 + h];
            if (c1[i * 9 + j] + c1[l * 9 + h] == b[i * 9 + j].count())
              dp[i * 9 + j][l * 9 + h] = 1;
            b[i * 9 + j] ^= b[l * 9 + h];
          }
        }
      }
    }
    int ctr1[10];
    long long ans = visit(ctr1, k, 0);
    cout << ans << "\n";
  }
  return 0;
}
