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 mod = 233333;
const long long N = 2e5 + 5;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long powmod(long long a, long long b) {
  long long ans = 1;
  a %= mod;
  while (b > 0) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans % mod;
}
long long readn() {
  long long sum = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    sum = sum * 10 + (ch - '0');
    ch = getchar();
  }
  return sum * f;
}
long long n, m, k, s, res, op, pw[N], sum[13][13];
vector<long long> v[13];
vector<pair<long long, long long> > G[N];
void dfs(long long x) {
  if (x == k + 1) {
    res += (op == s);
    return;
  }
  for (long long i = 0; i < x; i++) {
    op += sum[x][i];
    dfs(x + 1);
    op -= sum[x][i];
  }
}
int main() {
  n = readn();
  m = readn();
  k = readn();
  pw[0] = 1;
  for (long long i = 1; i <= n; i++) pw[i] = pw[i - 1] * mod, s += pw[i];
  for (long long i = 1; i <= m; i++) {
    long long u = readn(), v = readn(), w = readn();
    G[u].push_back({w, v});
  }
  for (long long i = 1; i <= n; i++)
    sort(G[i].begin(), G[i].end()), v[G[i].size()].push_back(i);
  for (long long i = 1; i <= k; i++) {
    for (long long j = 0; j < i; j++) {
      long long now = 0;
      for (long long p = 0; p < v[i].size(); p++) {
        now += pw[G[v[i][p]][j].second];
        sum[i][j] = now;
      }
    }
  }
  dfs(1);
  cout << res << endl;
}
