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;
struct node {
  int x, y;
  node(int nx = 0, int ny = 0) {
    x = nx;
    y = ny;
  }
} h;
int n, m, K, ot[200005], ans = 0, u, v, w, ba[10][10][10], vis[200005][10];
int ok[10], pr[10], gy[10];
vector<node> s[200005];
bool cmp(node a, node b) { return a.y < b.y; }
void dfs(int p) {
  if (p == K + 1) {
    for (int i = 1; i <= K; i++) ok[i] = gy[i];
    for (int i = 1; i < K; i++) {
      for (int j = i + 1; j <= K; j++) {
        ok[j] |= ba[i][pr[i]][j];
      }
    }
    for (int i = 1; i <= K; i++) {
      if ((1 << (pr[i] - 1)) & ok[i]) {
        return;
      }
    }
    ans++;
    return;
  }
  for (int i = 1; i <= p; i++) {
    pr[p] = i;
    dfs(p + 1);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &w);
    s[u].push_back(node(v, w));
    ot[u]++;
  }
  for (int i = 1; i <= n; i++) {
    sort(s[i].begin(), s[i].end(), cmp);
    for (int j = 0; j < s[i].size(); j++) {
      h = s[i][j];
      if (vis[h.x][ot[i]] & (1 << j))
        gy[ot[i]] |= (1 << j);
      else
        vis[h.x][ot[i]] |= (1 << j);
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= K; j++) {
      for (int k = 1; k <= j; k++) {
        if ((1 << (k - 1)) & vis[i][j]) {
          for (int u = j + 1; u <= K; u++) {
            ba[j][k][u] |= vis[i][u];
          }
        }
      }
    }
  }
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
