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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
const int N = 2e5 + 5;
const int HS = 3;
const int K = 10;
int n, m, k;
vector<pair<int, int>> adj[N];
int mod[HS];
int ans;
struct hash_num {
  int vals[HS];
  hash_num() { memset(vals, 0, sizeof(vals)); }
  hash_num(long long x) {
    for (int i = 0; i < HS; i++) {
      vals[i] = (x % mod[i] + mod[i]) % mod[i];
    }
  }
  hash_num operator+(hash_num x) {
    hash_num res = hash_num(0);
    for (int i = 0; i < HS - 1; i++) {
      res.vals[i] = (vals[i] + x.vals[i]) % mod[i];
    }
    res.vals[HS - 1] = (1LL * vals[HS - 1] * x.vals[HS - 1]) % mod[HS - 1];
    return res;
  }
  bool operator==(hash_num x) {
    for (int i = 0; i < HS; i++) {
      if (vals[i] != x.vals[i]) return false;
    }
    return true;
  }
} dp[K][K], values[N], sum;
void init() {
  uniform_int_distribution<int> rg(1e8, 1e9);
  for (int i = 0; i < HS; i++) {
    mod[i] = rg(rng32);
  }
}
void dfs(int u, hash_num x) {
  if (u == k) {
    if (x == sum) ++ans;
    return;
  }
  for (int v = 1; v <= u + 1; v++) {
    dfs(u + 1, x + dp[u + 1][v]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  init();
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    adj[u].emplace_back(w, v);
  }
  uniform_int_distribution<long long> rg(1, 1e18);
  for (int i = 0; i < n; i++) {
    values[i] = hash_num(rg(rng32));
    sum = sum + values[i];
  }
  for (int i = 0; i < n; i++) {
    sort(adj[i].begin(), adj[i].end());
    int len = adj[i].size();
    for (int j = 1; j <= len; j++) {
      dp[len][j] = dp[len][j] + values[adj[i][j - 1].second];
    }
  }
  dfs(0, hash_num(0));
  cout << ans << '\n';
  return 0;
}
