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;
struct TreeNode {
  int val;
  TreeNode* left;
  TreeNode* right;
  TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
struct ListNode {
  int val;
  ListNode* next;
  ListNode(int x) : val(x), next(NULL) {}
};
int nxt() {
  int x;
  cin >> x;
  return x;
}
long long nxtLong() {
  long long x;
  cin >> x;
  return x;
}
string nxtStr() {
  string x;
  cin >> x;
  return x;
}
char nxtChar() {
  char x;
  cin >> x;
  return x;
}
double nxtDouble() {
  double x;
  cin >> x;
  return x;
}
long double nxtLDouble() {
  long double x;
  cin >> x;
  return x;
}
TreeNode* makeTree(int n) {
  queue<TreeNode*> tree_nodes;
  TreeNode* root;
  if (n) {
    root = new TreeNode(nxt());
    tree_nodes.push(root);
  } else {
    return NULL;
  }
  while (!tree_nodes.empty()) {
    TreeNode* node = tree_nodes.front();
    tree_nodes.pop();
    bool left;
    cin >> left;
    if (left) {
      node->left = new TreeNode(nxt());
      tree_nodes.push(node->left);
    }
    bool right;
    cin >> right;
    if (right) {
      node->right = new TreeNode(nxt());
      tree_nodes.push(node->right);
    }
  }
  return root;
}
ListNode* makeLinkedList(int n) {
  ListNode* list = new ListNode(0);
  ListNode* dummy = list;
  for (int i = 0; i < n; ++i) {
    list->next = new ListNode(nxt());
    list = list->next;
  }
  return dummy->next;
}
const int mod = 1000000007;
void solve() {
  int n = nxt(), m = nxt(), k = nxt();
  vector<long long> differ(n + 1);
  differ[0] = 1ll;
  for (int i = 1; i <= n; ++i) {
    differ[i] = (differ[i - 1] * 10 % mod);
  }
  long long al = 0;
  for (int i = 1; i <= n; ++i) {
    al = (al + i * differ[i] % mod) % mod;
  }
  vector<vector<pair<int, int>>> graph(n + 1);
  for (int i = 0; i < m; ++i) {
    int u = nxt(), v = nxt(), wt = nxt();
    graph[u].push_back({wt, v});
  }
  vector<vector<vector<long long>>> vc(n + 1, vector<vector<long long>>(k + 1));
  vector<vector<long long>> a(k + 1, vector<long long>(k + 1));
  for (int i = 1; i <= n; ++i) {
    sort((graph[i]).begin(), (graph[i]).end());
    for (int j = 1; j <= graph[i].size(); ++j) {
      vc[graph[i].size()][j].push_back(graph[i][j - 1].second);
    }
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= i; ++j) {
      for (int u : vc[i][j]) {
        a[i][j] = (a[i][j] + u * differ[u] % mod) % mod;
      }
    }
  }
  long long cnt = 0;
  vector<int> t(k + 1);
  function<void(int)> dfs = [&](int u) {
    if (u == k + 1) {
      long long tmp = 0ll;
      for (int i = 1; i <= k; ++i) {
        tmp = (tmp + a[i][t[i]]) % mod;
      }
      if (tmp == al) {
        cnt++;
      }
      return;
    }
    for (int i = 1; i <= u; ++i) {
      t[u] = i;
      dfs(u + 1);
    }
  };
  dfs(1);
  cout << cnt << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(9);
  solve();
  return 0;
}
