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;
int n, m, k, chu[200010];
int hah[10][10][10][10], haha[10][10];
int st[10], ans, vis[200010];
vector<int> g[10];
vector<pair<int, int> > f1[200010];
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
bool pd() {
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= k; j++) {
      if (i != j) {
        if (hah[i][st[i]][j][st[j]] || haha[i][st[i]] || haha[j][st[j]])
          return false;
      }
    }
  }
  if (k == 1)
    if (haha[1][1]) return false;
  return true;
}
void dfs(int now) {
  if (now > k) {
    if (pd()) ans++;
    return;
  }
  for (int i = 1; i <= now; i++) {
    st[now] = i;
    dfs(now + 1);
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    chu[u]++;
    f1[u].push_back(make_pair(v, w));
  }
  for (int i = 1; i <= n; i++) sort(f1[i].begin(), f1[i].end(), cmp);
  for (int i = 1; i <= n; i++) g[chu[i]].push_back(i);
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int x = 1; x <= k; x++) {
        for (int y = 1; y <= x; y++) {
          for (unsigned l = 0; l < g[i].size(); l++) {
            int now = g[i][l];
            vis[f1[now][j - 1].first]++;
          }
          if (i != x) {
            for (unsigned l = 0; l < g[x].size(); l++) {
              int now = g[x][l];
              vis[f1[now][y - 1].first]++;
            }
          }
          for (int l = 1; l <= n; l++) {
            if (vis[l] > 1) {
              if (i == x && j == y)
                haha[i][j] = 1;
              else
                hah[i][j][x][y] = 1;
            }
            vis[l] = 0;
          }
        }
      }
    }
  }
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
