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 long long base = 200111;
const long long mod = 19260817;
vector<pair<int, int>> g[200111];
long long sm = 0;
long long pw[200111];
long long in[10][10], out[10][10];
int n, m, k, ans = 0;
vector<int> curr;
void dfs(int idx) {
  int i;
  if (idx == k) {
    long long cid = 0, cod = 0;
    for (i = 0; i < k; ++i) {
      cid += in[i][curr[i]];
      cid %= mod;
      cod += out[i][curr[i]];
      cod %= mod;
    }
    if (cid == sm && cod == sm) ans++;
    return;
  }
  for (i = 0; i <= idx; ++i) {
    curr[idx] = i;
    dfs(idx + 1);
  }
}
int main() {
  int i, j;
  scanf("%d%d%d", &n, &m, &k);
  pw[0] = 1;
  curr.resize(k);
  for (i = 1; i < 200111; ++i) {
    pw[i] = pw[i - 1] * base;
    pw[i] %= mod;
  }
  for (i = 1; i <= m; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    g[u].emplace_back(w, v);
  }
  for (i = 1; i <= n; ++i) {
    sort(g[i].begin(), g[i].end());
    int sz = g[i].size();
    for (j = 0; j < sz; ++j) {
      in[sz - 1][j] += pw[i];
      out[sz - 1][j] += pw[g[i][j].second];
      in[sz - 1][j] %= mod;
      out[sz - 1][j] %= mod;
    }
    sm += pw[i];
    sm %= mod;
  }
  dfs(0);
  printf("%d\n", ans);
  return 0;
}
