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;
using ll = long long;
using ld = double;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
ll hsh[200000];
vector<vector<pair<ll, ll>>> g;
ll n, m, k;
ll c[10];
ll res;
ll allVerticesHash = 1;
ll chash[10][10];
void rec(int i) {
  if (i == k) {
    ll cur = 1;
    for (ll j = 0; j < (ll)k; ++j) {
      cur += chash[j][c[j]];
      cur %= ((ll)(1e9 + 7));
    }
    res += cur == allVerticesHash;
    return;
  }
  for (ll j = 0; j < (ll)i + 1; ++j) {
    c[i] = j;
    rec(i + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  srand(time(0));
  cin >> n >> m >> k;
  for (ll i = 0; i < (ll)n; ++i) {
    hsh[i] = rand() + 2;
    allVerticesHash += hsh[i];
    allVerticesHash %= ((ll)(1e9 + 7));
  }
  g = vector<vector<pair<ll, ll>>>(n);
  for (ll i = 0; i < (ll)m; ++i) {
    ll u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    g[u].push_back({w, v});
  }
  for (ll i = 0; i < (ll)n; ++i) sort(g[i].begin(), g[i].end());
  for (ll i = 0; i < (ll)k; ++i) {
    for (int v = 0; v < n; ++v) {
      if (g[v].size() != i + 1) continue;
      for (ll j = 0; j < (ll)i + 1; ++j) {
        chash[i][j] += hsh[g[v][j].second];
        chash[i][j] %= ((ll)(1e9 + 7));
      }
    }
  }
  rec(0);
  cout << res << '\n';
  return 0;
}
