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 long long infl = 0x3f3f3f3f3f3f3f3fLL;
const long long infi = 0x3f3f3f3f;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 1e9 + 7;
const long long N = 2e5 + 5;
long long n, m, k;
vector<pair<long long, long long>> g[N];
vector<pair<long long, long long>> in[N];
long long val[10][10];
long long hs[N];
long long s = 0, ans = 0;
void dfs(long long x, long long h) {
  if (x == k) {
    ans += (h == s);
    return;
  }
  for (long long i = 1; i <= x + 1; i++) dfs(x + 1, (h + val[x + 1][i]) % mod);
}
void solve() {
  uniform_int_distribution<long long> rg(0, 1e9);
  cin >> n >> m >> k;
  for (long long i = 1; i <= m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    g[u].push_back({w, v});
  }
  for (long long i = 1; i <= n; i++) {
    hs[i] = rg(mt);
    s = (s + hs[i]) % mod;
  }
  for (long long i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    for (long long j = 0; j < ((long long)g[i].size()); j++) {
      val[((long long)g[i].size())][j + 1] += hs[g[i][j].second];
      val[((long long)g[i].size())][j + 1] %= mod;
    }
  }
  dfs(0, 0);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  long long t = 1;
  while (t--) solve();
  return 0;
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % mod) * powm(b, mod - 2) % mod;
}
long long norm(long long a) {
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
  return a;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
