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 solprod[10][10];
long long solsum[10][10];
const int MOD = 1e9 + 7;
int rezprodmod = 1, cnt = 0, k;
long long rezsummod = 1;
vector<pair<int, int>> v[200001];
void bkt(int poz, int rezprod, long long rezsum) {
  if (poz == k + 1) {
    if (rezprod == rezprodmod && rezsum == rezsummod) cnt++;
    return;
  }
  for (int i = 1; i <= poz; i++)
    bkt(poz + 1, 1LL * rezprod * solprod[poz][i] % MOD,
        rezsum + solsum[poz][i]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, i, a, b, c, j;
  cin >> n >> m >> k;
  for (i = 1; i <= m; i++) {
    cin >> a >> b >> c;
    v[a].push_back({c, b});
  }
  for (i = 1; i <= 9; i++)
    for (j = 1; j <= 9; j++) {
      solprod[i][j] = 1;
      solsum[i][j] = 0;
    }
  for (i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  for (i = 1; i <= n; i++)
    for (j = 0; j < v[i].size(); j++) {
      solprod[v[i].size()][j + 1] =
          1LL * solprod[v[i].size()][j + 1] * v[i][j].second % MOD;
      solsum[v[i].size()][j + 1] = solsum[v[i].size()][j + 1] + v[i][j].second;
    }
  for (i = 1; i <= n; i++) rezprodmod = 1LL * rezprodmod * i % MOD;
  rezsummod = 1LL * n * (n + 1) / 2;
  bkt(1, 1, 0);
  cout << cnt;
  return 0;
}
