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 int maxl = 3e5 + 10;
int n, m, cas, k, cnt, tot, ans;
int a[maxl], out[maxl], c[maxl];
struct ed {
  int u, v, l;
};
vector<ed> e[maxl];
long long in[100][100];
int num[100][100];
set<pair<int, int> > s[10][10];
vector<pair<int, int> > b[maxl];
inline bool cmp(const ed &a, const ed &b) { return a.l < b.l; }
inline void prework() {
  scanf("%d%d%d", &n, &m, &k);
  int u, v, l;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &l);
    out[u]++;
    e[u].push_back(ed{u, v, l});
  }
  for (int i = 1; i <= n; i++) sort(e[i].begin(), e[i].end(), cmp);
  for (int i = 1; i <= n; i++) {
    l = e[i].size();
    for (int j = 0; j < l; j++) b[e[i][j].v].push_back({out[e[i][j].u], j + 1});
  }
}
inline void dfs(int x, int res) {
  if (x > k) {
    if (res == 0) ans++;
    return;
  }
  if (c[x] > 0) {
    if (!in[x][c[x]]) {
      a[x] = c[x];
      for (pair<int, int> d : s[x][c[x]]) ++in[d.first][d.second];
      dfs(x + 1, res - num[x][c[x]]);
      a[x] = 0;
      for (pair<int, int> d : s[x][c[x]]) --in[d.first][d.second];
    }
  } else {
    for (int i = 1; i <= x; i++)
      if (!in[x][i]) {
        a[x] = i;
        for (pair<int, int> d : s[x][i]) ++in[d.first][d.second];
        dfs(x + 1, res - num[x][i]);
        for (pair<int, int> d : s[x][i]) --in[d.first][d.second];
        a[x] = 0;
      }
  }
}
inline void mainwork() {
  int u, v, l;
  ans = 0;
  for (int i = 1; i <= n; i++) {
    l = b[i].size();
    if (l == 0)
      return;
    else if (l == 1) {
      u = b[i][0].first;
      num[u][b[i][0].second]++;
      if (c[u] == 0)
        c[b[i][0].first] = b[i][0].second;
      else if (c[u] != b[i][0].second)
        return;
    } else {
      sort(b[i].begin(), b[i].end());
      for (int j = 0; j < l; j++)
        if (j > 1 && b[i][j] == b[i][j - 1])
          in[b[i][j].first][b[i][j].second]++;
      b[i].erase(unique(b[i].begin(), b[i].end()), b[i].end());
      l = b[i].size();
      for (int j = 0; j < l; j++) {
        num[b[i][j].first][b[i][j].second]++;
        for (int k = 0; k < l; k++)
          if (j != k) {
            u = b[i][j].first;
            v = b[i][k].first;
            s[u][b[i][j].second].insert(b[i][k]);
            s[v][b[i][k].second].insert(b[i][j]);
          }
      }
    }
  }
  dfs(1, n);
}
inline void print() { printf("%d\n", ans); }
int main() {
  int t = 1;
  for (cas = 1; cas <= t; cas++) {
    prework();
    mainwork();
    print();
  }
  return 0;
}
