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>
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3", "omit-frame-pointer", "inline")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2")
using namespace std;
const long long mod = 1000000007;
const long long mod2 = 998244353;
mt19937 rng((int)chrono::steady_clock::now().time_since_epoch().count());
const int mxN = 200010;
const int mxK = 10;
long long nxt[2][mxK][mxK];
long long sum[2];
long long val[2][mxN];
int k;
int backtrack(int at, long long cur1, long long cur2) {
  if (at > k) return cur1 == sum[0] && cur2 == sum[1];
  int res = 0;
  for (int i = 0; i < at; i++)
    res += backtrack(at + 1, cur1 * nxt[0][at][i] % mod,
                     cur2 * nxt[1][at][i] % mod2);
  return res;
}
vector<pair<int, int> > g[mxN];
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  srand(time(NULL));
  cout << fixed << setprecision(9);
  int t = 1;
  while (t--) {
    int n, m;
    cin >> n >> m >> k;
    sum[0] = sum[1] = 1;
    for (int u = 0; u < n; u++) {
      val[0][u] = rng() % mod;
      sum[0] = sum[0] * val[0][u] % mod;
      val[1][u] = rng() % mod2;
      sum[1] = sum[1] * val[1][u] % mod2;
    }
    for (int i = 0, u, v, w; i < m; i++) {
      cin >> u >> v >> w;
      u--, v--;
      g[u].push_back({w, v});
    }
    for (int i = 1; i <= k; i++)
      for (int j = 0; j < i; j++) nxt[0][i][j] = nxt[1][i][j] = 1;
    for (int u = 0; u < n; u++) {
      sort(begin(g[u]), end(g[u]));
      for (int i = 0; i < (int)g[u].size(); i++) {
        auto v = g[u][i].second;
        nxt[0][g[u].size()][i] = (nxt[0][g[u].size()][i] * val[0][v]) % mod;
        nxt[1][g[u].size()][i] = (nxt[1][g[u].size()][i] * val[1][v]) % mod2;
      }
    }
    cout << backtrack(1, 1, 1) << '\n';
  }
  return 0;
}
