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 N = 2e5 + 10;
const int mod = 1e9 + 7;
const int M = 10;
int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + c - '0', c = getchar();
  return x * f;
}
int sum[M][M][M][M];
int e[N][M][M];
vector<pair<int, int> > g[N];
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int c[M];
int k, ans;
void dfs(int x) {
  if (x == k + 1) {
    int res = 1;
    for (int i = 1; i <= k; i++)
      if (sum[i][c[i]][i][c[i]]) {
        res = 0;
        break;
      }
    for (int i = 1; i <= k; i++)
      for (int j = i + 1; j <= k; j++)
        if (sum[i][c[i]][j][c[j]]) {
          res = 0;
          break;
        }
    ans += res;
    return;
  }
  for (int i = 1; i <= x; i++) {
    c[x] = i;
    dfs(x + 1);
  }
}
int main() {
  int n = read(), m = read();
  k = read();
  for (int i = 1; i <= m; i++) {
    int u = read(), v = read(), w = read();
    g[u].push_back(make_pair(v, w));
  }
  for (int i = 1; i <= n; i++) sort(g[i].begin(), g[i].end(), cmp);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < g[i].size(); j++) {
      int to = g[i][j].first;
      e[to][g[i].size()][j + 1]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int x1 = 1; x1 <= k; x1++) {
      for (int y1 = 1; y1 <= x1; y1++) {
        if (e[i][x1][y1]) {
          for (int x2 = 1; x2 <= k; x2++) {
            for (int y2 = 1; y2 <= x2; y2++) {
              if (x1 == x2 && y1 == y2) continue;
              if (e[i][x2][y2]) sum[x1][y1][x2][y2] = sum[x2][y2][x1][y1] = 1;
            }
          }
        }
        if (e[i][x1][y1] >= 2) sum[x1][y1][x1][y1] = 1;
      }
    }
  }
  dfs(1);
  cout << ans << endl;
}
