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;
mt19937 mt_rand(
    chrono::high_resolution_clock::now().time_since_epoch().count());
const int maxn = 2e5 + 5;
vector<pair<int, int>> v[maxn];
bool cmp(pair<int, int> a, pair<int, int> b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second < b.second;
}
long long w[10][10];
int mod[3];
int n, m, k;
struct hash_number {
  int hs = 3;
  long long a[3];
  hash_number() { fill(a, a + hs, 0); }
  hash_number(long long x) {
    for (int i = 1; i < hs; i++) {
      a[i] = (x % mod[i] + mod[i]) % mod[i];
    }
  }
  hash_number operator+(hash_number x) {
    hash_number res;
    for (int i = 0; i < 3; i++) {
      res.a[i] = (a[i] + x.a[i]) % mod[i];
    }
    return res;
  }
  bool operator==(const hash_number &x) const {
    for (int i = 0; i < hs; i++) {
      if (a[i] != x.a[i]) return false;
    }
    return true;
  }
} val[maxn], c[10][10], s;
int dfs(int x, hash_number now) {
  if (x == k + 1) {
    return now == s;
  }
  int ans = 0;
  for (int i = 1; i <= x; i++) {
    ans += dfs(x + 1, now + c[x][i]);
  }
  return ans;
}
int main() {
  mod[0] = 998244353;
  mod[1] = 1e9 + 7;
  mod[2] = uniform_int_distribution<int>(1e8, 1e9)(mt_rand);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, to, w;
    cin >> u >> to >> w;
    v[u].push_back(make_pair(to, w));
  }
  uniform_int_distribution<long long> rg(1, 1e18);
  for (int i = 1; i <= n; i++) {
    val[i] = hash_number(rg(mt_rand));
  }
  for (int i = 1; i <= n; i++) s = s + val[i];
  for (int i = 1; i <= n; i++) {
    int d = v[i].size();
    sort(v[i].begin(), v[i].end(), cmp);
    for (int j = 0; j < v[i].size(); j++) {
      c[d][j + 1] = c[d][j + 1] + val[v[i][j].first];
    }
  }
  cout << dfs(1, hash_number()) << endl;
  return 0;
}
