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 BASE = 10;
const int MOD[3] = {1000000007, 1000000009, 1000000021};
const int MAXN = 2E5;
int target[3];
vector<pair<int, int>> adjList[MAXN + 1];
int res[10][10][3];
bool visited[MAXN + 1];
int conf[10];
int solve(int ind, int k) {
  if (ind > k) {
    for (int i = 0; i < 3; i++) {
      int temp = 0;
      for (int j = 1; j <= k; j++) {
        temp += res[j][conf[j]][i];
        if (temp >= MOD[i]) temp -= MOD[i];
      }
      if (temp != target[i]) return 0;
    }
    return 1;
  }
  int ans = 0;
  for (int i = 0; i < ind; i++) {
    if (res[ind][i][0] == -1) continue;
    conf[ind] = i;
    ans += solve(ind + 1, k);
  }
  return ans;
}
int getVal(int n, int mod) {
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = (int)((long long)ans * BASE % mod);
    ans += visited[i];
    if (ans >= mod) ans -= mod;
  }
  return ans;
}
int calc(int ind, int val, int n, int mod) {
  fill(visited + 1, visited + n + 1, false);
  for (int i = 1; i <= n; i++) {
    if ((int)adjList[i].size() != ind) continue;
    int v = adjList[i][val].second;
    if (visited[v]) return -1;
    visited[v] = true;
  }
  return getVal(n, mod);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adjList[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    sort(adjList[i].begin(), adjList[i].end());
  }
  fill(visited + 1, visited + n + 1, true);
  for (int i = 0; i < 3; i++) {
    target[i] = getVal(n, MOD[i]);
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < i; j++) {
      for (int modInd = 0; modInd < 3; modInd++) {
        res[i][j][modInd] = calc(i, j, n, MOD[modInd]);
      }
    }
  }
  int ans = solve(1, k);
  cout << ans << "\n";
  return 0;
}
