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;
long long M[2];
struct str {
  long long r[2];
  str(long long x = 0) {
    for (long long i = 0; i < 2; ++i) r[i] = x;
  }
  str& operator+=(str b) {
    for (long long i = 0; i < 2; ++i) r[i] = (r[i] + b.r[i]) % M[i];
    return *this;
  }
  friend str operator+(str a, str b) { return a += b; }
  str& operator-=(str b) {
    for (long long i = 0; i < 2; ++i) r[i] = (r[i] - b.r[i] + M[i]) % M[i];
    return *this;
  }
  friend str operator-(str a, str b) { return a -= b; }
  str& operator*=(str b) {
    for (long long i = 0; i < 2; ++i) r[i] = (1LL * r[i] * b.r[i]) % M[i];
    return *this;
  }
  friend str operator*(str a, str b) { return a *= b; }
};
bool operator==(str a, str b) {
  for (long long i = 0; i < 2; ++i)
    if (a.r[i] != b.r[i]) return false;
  return true;
}
long long n, m, k;
str second[9][9];
long long ans = 0;
str a = 0;
void gen(long long pos, str now) {
  if (pos == k) {
    if (now == a) ans++;
  } else {
    for (long long i = 0; i <= pos; ++i) gen(pos + 1, now + second[pos][i]);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  M[0] = 1000000007;
  M[1] = 1000000009;
  cin >> n >> m >> k;
  vector<vector<pair<long long, long long>>> g(n);
  for (long long i = 0; i < m; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    g[u].push_back({w, v});
  }
  for (long long i = 0; i < n; ++i) sort(g[i].begin(), g[i].end());
  vector<str> val(n);
  for (long long i = 0; i < n; ++i) val[i] = rand();
  for (long long i = 0; i < k; ++i) {
    for (long long j = 0; j <= i; ++j) {
      for (long long v = 0; v < n; ++v) {
        if (g[v].size() == i + 1)
          second[i][j] = second[i][j] + val[g[v][j].second];
      }
    }
  }
  for (long long i = 0; i < n; ++i) a += val[i];
  gen(0, 0);
  cout << ans;
  return 0;
}
