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;
set<pair<long long, long long> > tu[200005];
long long base = 233333;
long long mod1 = 1e9 + 7;
long long mod2 = 104857601;
long long n, m, k;
long long quick_mul(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    res %= mod;
    a = a * a % mod;
    a %= mod;
    b >>= 1;
  }
  return res % mod;
}
long long zong1 = 0;
long long zong2 = 0;
long long jilu1[10][10];
long long jilu2[10][10];
long long res = 0;
void dfs(long long lim, long long now1, long long now2) {
  if (lim == k + 1) {
    if (now1 % mod1 == zong1 && now2 % mod2 == zong2) res++;
    return;
  }
  for (int i = 1; i <= lim; i++) {
    dfs(lim + 1, (now1 + jilu1[lim][i]) % mod1, (now2 + jilu2[lim][i]) % mod2);
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    tu[u].insert(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    int len = tu[i].size();
    int temp = 1;
    for (auto j : tu[i]) {
      jilu1[len][temp] += quick_mul(base, j.second, mod1);
      jilu1[len][temp] %= mod1;
      jilu2[len][temp] += quick_mul(base, j.second, mod2);
      jilu2[len][temp] %= mod2;
      temp++;
    }
  }
  for (int i = 1; i <= n; i++) {
    zong1 += quick_mul(base, i, mod1);
    zong1 %= mod1;
    zong2 += quick_mul(base, i, mod2);
    zong2 %= mod2;
  }
  dfs(1, 0, 0);
  cout << res << endl;
  return 0;
}
