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;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void ran() { srand(chrono::steady_clock::now().time_since_epoch().count()); }
long long get_rand() {
  long long a = rand();
  long long b = rand();
  return a * (RAND_MAX + 1ll) + b;
}
void usaco() {
  freopen("problem.in", "r", stdin);
  freopen("problem.out", "w", stdout);
}
int n, m, k;
const int MXN = 200005;
vector<pair<int, int>> inc[MXN], out[MXN];
set<pair<pair<int, int>, pair<int, int>>> tog;
vector<pair<pair<int, int>, pair<int, int>>> tl;
vector<pair<int, int>> cr;
int co = 0;
int og[10][10], on[10][10];
int alr, exi;
int pm, px;
int fn(int x) { return x * pm + x * px + x * x * 4 + 1; }
int fg(int x) { return x * px + x * pm * 4 + 19; }
int fng(int x) { return fn(fg(x)); }
void dfs(int at) {
  if (at > k) {
    for (int i = 0; i < cr.size(); i++) {
      for (int j = i + 1; j < cr.size(); j++) {
        pair<pair<int, int>, pair<int, int>> t = {cr[i], cr[j]};
        auto te = lower_bound(tl.begin(), tl.end(), t);
        if (te != tl.end() && *te == t) {
          return;
        }
      }
    }
    int sm = 0, sn = 0;
    for (auto x : cr) {
      sm += og[x.first][x.second];
      sn += on[x.first][x.second];
    }
    if (sm != alr || sn != exi) return;
    co++;
    return;
  }
  for (int i = 1; i <= at; i++) {
    cr.push_back({at, i});
    dfs(at + 1);
    cr.pop_back();
  }
}
signed main() {
  ran();
  fast();
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    out[a].push_back({w, b});
    inc[b].push_back({w, a});
  }
  for (int i = 1; i <= n; i++) {
    sort(inc[i].begin(), inc[i].end());
    sort(out[i].begin(), out[i].end());
  }
  vector<pair<int, int>> poss;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      poss.push_back({i, j});
    }
  }
  for (int i = 1; i <= n; i++) {
    alr += fn(i);
    exi += fg(i);
    exi += fng(i);
    int o = out[i].size();
    for (int j = 0; j < o; j++) {
      og[o][j + 1] += (fn(out[i][j].second));
      on[o][j + 1] += fg(out[i][j].second);
      on[o][j + 1] += fng(out[i][j].second);
    }
  }
  for (int i = 1; i <= n; i++) {
    set<pair<int, int>> st;
    for (auto x : inc[i]) {
      int from = x.second;
      int with = x.first;
      pair<int, int> se = {with, i};
      int of = lower_bound(out[from].begin(), out[from].end(), se) -
               out[from].begin() + 1;
      int de = out[from].size();
      st.insert({de, of});
    }
    vector<pair<int, int>> tr(st.begin(), st.end());
    for (int i = 0; i < tr.size(); i++) {
      for (int j = i + 1; j < tr.size(); j++) {
        tog.insert({min(tr[i], tr[j]), max(tr[i], tr[j])});
      }
    }
  }
  tl = vector<pair<pair<int, int>, pair<int, int>>>(tog.begin(), tog.end());
  dfs(1);
  cout << co << endl;
}
