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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long n, m, k, rr[200005];
vector<pair<long long, long long> > vr[200005], tr[200005];
long long add[10][10], inv[10][10][10][10], qr[10], er[10];
long long func(long long, long long), tot = 0;
int main() {
  unordered_set<long long> sr;
  scanf("%lld %lld %lld", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    long long x, y, z;
    scanf("%lld %lld %lld", &x, &y, &z);
    vr[x].push_back(make_pair(z, y));
    tr[y].push_back(make_pair(z, x));
  }
  while (sr.size() != n) sr.insert(rng() % 10000000000000ll);
  long long ptr = 1;
  for (auto itr : sr) rr[ptr++] = (itr);
  long long ok = 1;
  for (int i = 1; i < n + 1; i++) {
    er[vr[i].size()] = 1;
    if (vr[i].size() > 9) ok = 0;
    if (tr[i].size() == 0) ok = 0;
    sort(vr[i].begin(), vr[i].end());
  }
  if (!ok) {
    printf("0");
    return 0;
  }
  for (int i = 1; i < n + 1; i++) {
    tot += rr[i];
    for (int j = 0; j < vr[i].size(); j++) {
      add[vr[i].size()][j] += rr[vr[i][j].second];
    }
  }
  printf("%lld\n", func(1, 0));
}
long long func(long long idx, long long ans) {
  if (idx == k + 1) {
    return ans == tot;
  }
  long long aa = 0;
  for (int i = 0; i < idx; i++) {
    aa += func(idx + 1, ans + add[idx][i]);
  }
  return aa;
}
