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 maxi = 4e5 + 10;
int a[maxi];
string s;
int n, m, k;
vector<pair<int, int> > v[maxi];
int ans;
bool c[10][10][10][10];
int deg[maxi];
int d[maxi][10][10];
bool nope[10][10];
int cur[10];
bool cmp(pair<int, int> x, pair<int, int> y) { return x.second < y.second; }
void rec(int x) {
  if (x == k + 1) {
    ans++;
    return;
  }
  for (int i = 1; i <= x; i++) {
    int ok = 1;
    if (nope[x][i]) ok = 0;
    for (int j = x - 1; j > 0 && ok; j--)
      if (c[x][i][j][cur[j]]) {
        ok = 0;
        break;
      }
    if (ok) {
      cur[x] = i;
      rec(x + 1);
    }
  }
}
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    deg[x]++;
    v[x].push_back({y, z});
  }
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end(), cmp);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < deg[i]; j++) d[v[i][j].first][deg[i]][j + 1]++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= k; j++)
      for (int l = 1; l <= j; l++) {
        if (d[i][j][l] > 1) {
          nope[j][l] = 1;
          continue;
        }
        if (!d[i][j][l]) continue;
        for (int o = 1; o < j; o++)
          for (int u = 1; u <= o; u++)
            if (d[i][o][u]) {
              c[j][l][o][u] = 1;
            }
      }
  }
  rec(1);
  cout << ans << endl;
}
int main() {
  int t;
  t = 1;
  while (t--) solve();
}
