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;
int fac[15];
int ans[3700000];
vector<pair<int, int> > e[200010];
vector<pair<int, int> > qwq[200010];
int main() {
  scanf("%d%d%d", &n, &m, &k);
  fac[0] = 1;
  for (int i = 1; i <= k + 1; i++) fac[i] = i * fac[i - 1];
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    e[u].push_back({w, v});
  }
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end());
    for (int j = 0; j < e[i].size(); j++) {
      qwq[e[i][j].second].push_back({e[i].size(), j + 1});
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(qwq[i].begin(), qwq[i].end());
    for (int j = 1; j < qwq[i].size(); j++)
      if (qwq[i][j] == qwq[i][j - 1])
        ans[qwq[i][j].second * fac[qwq[i][j].first]] = 1;
    int cnt = unique(qwq[i].begin(), qwq[i].end()) - qwq[i].begin();
    for (int j = 0; j < cnt; j++)
      for (int k = j + 1; k < cnt; k++) {
        if (qwq[i][j].first == qwq[i][k].first) continue;
        ans[qwq[i][j].second * fac[qwq[i][j].first] +
            qwq[i][k].second * fac[qwq[i][k].first]] = 1;
      }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < fac[k + 1]; j++) {
      if (j / fac[i] % (i + 1) == 0) {
        for (int q = 1; q <= i; q++) ans[j + q * fac[i]] |= ans[j];
      }
    }
  }
  int res = 0;
  for (int i = 0; i < fac[k + 1]; i++) {
    int qwq = 1;
    for (int j = 1; j <= k; j++)
      if (i / fac[j] % (j + 1) == 0) qwq = 0;
    if (qwq == 1 && ans[i] == 0) res++;
  }
  printf("%d\n", res);
}
