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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int maxn = 2e5 + 5, mod = 1e9 + 7, inf = mod;
bool ban[10][10][10][10];
int n, m, k;
vector<pair<int, int> > g[maxn];
multiset<pair<int, int> > in[maxn];
int ans;
int b[maxn];
void back(int i) {
  if (i == k + 1) {
    for (int i1 = (int)1; i1 <= (int)k; i1++) {
      for (int i2 = (int)1; i2 <= (int)k; i2++) {
        if (ban[i1][b[i1]][i2][b[i2]]) {
          return;
        }
      }
    }
    ans++;
  } else {
    for (int j = (int)1; j <= (int)i; j++) {
      b[i] = j;
      back(i + 1);
    }
  }
}
bool cmp(pair<int, int> i, pair<int, int> j) { return i.second < j.second; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = (int)1; i <= (int)m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].emplace_back(v, w);
  }
  for (int i = (int)1; i <= (int)n; i++) {
    sort(begin(g[i]), end(g[i]), cmp);
    int t = g[i].size();
    for (int j = (int)0; j <= (int)t - 1; j++) {
      int v = g[i][j].first;
      if (in[v].count({t, j + 1}) > 1) continue;
      in[v].emplace(t, j + 1);
    }
  }
  for (int i = (int)1; i <= (int)n; i++) {
    for (pair<int, int> a : in[i])
      if (in[i].count(a) >= 2) {
        ban[a.first][a.second][a.first][a.second] = 1;
      }
    for (pair<int, int> a : in[i]) {
      for (pair<int, int> b : in[i]) {
        if (a == b) continue;
        ban[a.first][a.second][b.first][b.second] = 1;
      }
    }
  }
  back(1);
  cout << ans;
}
