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 maxN = 200005;
const int n_mods = 2;
const int mod[n_mods] = {1000000007, 1000000009};
mt19937 mt_rand(
    chrono::high_resolution_clock::now().time_since_epoch().count());
struct HashNumber {
  long long xs[n_mods];
  HashNumber() {
    for (int i = 0; i < n_mods; ++i) xs[i] = 0;
  }
  HashNumber(long long x) {
    for (int i = 0; i < n_mods; ++i) xs[i] = x % mod[i];
  }
  HashNumber operator+(HashNumber other) {
    HashNumber x;
    for (int i = 0; i < n_mods; ++i) x.xs[i] = (xs[i] + other.xs[i]) % mod[i];
    return x;
  }
  bool operator==(HashNumber other) {
    for (int i = 0; i < n_mods; ++i) {
      if (xs[i] != other.xs[i]) return false;
    }
    return true;
  }
};
int N, M, K;
vector<pair<int, int>> graph[maxN];
HashNumber target, S[10][10], val[maxN];
bool cmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int solve(int k, HashNumber s) {
  if (k == K) {
    if (s == target) return 1;
    return 0;
  }
  int ans = 0;
  for (int i = 0; i <= k; ++i) {
    ans += solve(k + 1, s + S[k + 1][i + 1]);
  }
  return ans;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> K;
  for (int i = 0; i < M; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    graph[u].push_back({v, w});
  }
  std::uniform_int_distribution<long long> interval(1, (long long)1e18);
  for (int i = 1; i <= N; ++i) val[i] = HashNumber(interval(mt_rand));
  for (int i = 1; i <= N; ++i) target = target + val[i];
  for (int i = 1; i <= N; ++i) {
    sort(graph[i].begin(), graph[i].end(), cmp);
    int sz = (int)graph[i].size();
    for (int j = 0; j < sz; ++j) {
      S[sz][j + 1] = S[sz][j + 1] + val[graph[i][j].first];
    }
  }
  printf("%d\n", solve(0, HashNumber()));
  return 0;
}
