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 maxn = 2e5 + 7;
struct Node {
  long long to, weight;
};
vector<Node> field[maxn];
vector<long long> sum[10];
vector<long long> hash1[10], hash2[10];
long long n, m, k;
long long ans = 0;
long long tar1, tar2;
bool cmp(Node a, Node b) { return a.weight < b.weight; }
void dfs(long long deep, long long tot1, long long tot2) {
  if (deep == k + 1) {
    if (tot1 == tar1 && tot2 == tar2) ans++;
    return;
  }
  for (long long i = 0; i < deep; i++) {
    dfs(deep + 1, tot1 + hash1[deep][i], tot2 + hash2[deep][i]);
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  for (long long i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    field[u].push_back({v, w});
  }
  for (long long i = 1; i <= n; i++) {
    tar1 += i * i * i;
    tar2 += i * i;
    sort(field[i].begin(), field[i].end(), cmp);
    sum[field[i].size()].push_back(i);
  }
  for (long long i = 1; i <= k; i++) {
    for (long long j = 0; j < i; j++) {
      unordered_map<long long, long long> M;
      long long temp1 = 0, temp2 = 0;
      for (auto &x : sum[i]) {
        long long to = field[x][j].to;
        temp1 += to * to * to;
        temp2 += to * to;
      }
      hash1[i].push_back(temp1);
      hash2[i].push_back(temp2);
    }
  }
  dfs(1, 0, 0);
  cout << ans << endl;
}
