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 N = 200010;
int n, m, k;
vector<pair<int, int> > g[N];
vector<pair<int, int> > arr[N];
const int K = 15;
bool bad[K][K][K][K];
vector<int> v;
int ans = 0;
void calc(int i) {
  if (i == (int)v.size()) {
    ans++;
    return;
  }
  bool good;
  for (int j = 0; j <= i; j++) {
    v[i] = j;
    good = true;
    for (int k = 0; k <= i; k++) {
      if (bad[i][j][k][v[k]]) {
        good = false;
        break;
      }
    }
    if (!good) continue;
    calc(i + 1);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  v.resize(k, 0);
  for (int a, b, c, i = 0; i < m; i++) {
    scanf("%d%d%d", &a, &b, &c);
    g[a].push_back(make_pair(c, b));
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    for (int j = 0; j < (int)g[i].size(); j++) {
      arr[g[i][j].second].push_back(make_pair((int)g[i].size() - 1, j));
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(arr[i].begin(), arr[i].end());
    for (int j = 0; j + 1 < (int)arr[i].size(); j++) {
      if (arr[i][j] == arr[i][j + 1])
        bad[arr[i][j].first][arr[i][j].second][arr[i][j].first]
           [arr[i][j].second] = true;
    }
    arr[i].resize(unique(arr[i].begin(), arr[i].end()) - arr[i].begin());
    for (int j = 0; j < (int)arr[i].size(); j++) {
      for (int k = 0; k < (int)arr[i].size(); k++) {
        if (j == k) continue;
        bad[arr[i][j].first][arr[i][j].second][arr[i][k].first]
           [arr[i][k].second] = true;
      }
    }
  }
  calc(0);
  cout << ans << endl;
  return 0;
}
