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 long long MOD = 998244353;
const long long BASE1 = 123456789;
const long long BASE2 = 987654321;
const int MAX = 200000;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  vector<long long> power1(MAX, 1);
  for (int i = 1; i < MAX; i++) {
    power1[i] = power1[i - 1] * BASE1 % MOD;
  }
  vector<long long> power2(MAX, 1);
  for (int i = 1; i < MAX; i++) {
    power2[i] = power2[i - 1] * BASE2 % MOD;
  }
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> E(n);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    E[u].push_back(make_pair(w, v));
  }
  for (int i = 0; i < n; i++) {
    sort(E[i].begin(), E[i].end());
  }
  vector<vector<int>> vl(k);
  for (int i = 0; i < n; i++) {
    vl[E[i].size() - 1].push_back(i);
  }
  vector<vector<pair<long long, long long>>> V(k);
  for (int i = 0; i < k; i++) {
    V[i] = vector<pair<long long, long long>>(i + 1);
    for (int j = 0; j <= i; j++) {
      long long h1 = 0, h2 = 0;
      for (int v : vl[i]) {
        int w = E[v][j].second;
        h1 += power1[w];
        h1 %= MOD;
        h2 += power2[w];
        h2 %= MOD;
      }
      V[i][j] = make_pair(h1, h2);
    }
  }
  map<pair<long long, long long>, int> dp;
  dp[make_pair(0, 0)] = 1;
  for (int i = 0; i < k; i++) {
    map<pair<long long, long long>, int> dp2;
    for (auto P : dp) {
      long long h1 = P.first.first;
      long long h2 = P.first.second;
      int cnt = P.second;
      for (int j = 0; j <= i; j++) {
        dp2[make_pair((h1 + V[i][j].first) % MOD,
                      (h2 + V[i][j].second) % MOD)] += cnt;
      }
    }
    swap(dp, dp2);
  }
  long long c1 = 0, c2 = 0;
  for (int i = 0; i < n; i++) {
    c1 += power1[i];
    c1 %= MOD;
    c2 += power2[i];
    c2 %= MOD;
  }
  cout << dp[make_pair(c1, c2)] << endl;
}
