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 maxn = 2e5 + 5;
vector<pair<int, int> > lik[maxn];
vector<int> out[maxn];
int rk[maxn][10];
int n, m, k;
int in[maxn];
int base1 = 31, base2 = 57;
unsigned long long ha1[10][10], ha2[10][10];
unsigned long long ph1[maxn], ph2[maxn];
unsigned long long tar1, tar2;
bool cmp(pair<int, int> lhs, pair<int, int> rhs) {
  return lhs.second < rhs.second;
}
void init() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; ++i) {
    ph1[i] = ph1[i - 1] * base1 + i;
    ph2[i] = ph2[i - 1] * base2 + i;
  }
  int u, v, w;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d%d", &u, &v, &w);
    lik[u].push_back(pair<int, int>(v, w));
  }
  for (int i = 1; i <= n; ++i) {
    sort(lik[i].begin(), lik[i].end(), cmp);
    for (int j = 0; j < lik[i].size(); ++j) rk[i][j + 1] = lik[i][j].first;
    out[lik[i].size()].push_back(i);
    tar1 += ph1[i];
    tar2 += ph2[i];
  }
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j)
      for (auto u : out[i])
        ha1[i][j] += ph1[rk[u][j]], ha2[i][j] += ph2[rk[u][j]];
}
int ans;
void dfs(int s, unsigned long long h1, unsigned long long h2) {
  if (s > k) {
    if (h1 == tar1 && h2 == tar2) ++ans;
    return;
  }
  for (int c = 1; c <= s; ++c) {
    dfs(s + 1, h1 + ha1[s][c], h2 + ha2[s][c]);
  }
}
void sol() {
  dfs(1, 0, 0);
  printf("%d\n", ans);
}
int main() {
  int T;
  {
    init();
    sol();
  }
  return 0;
}
