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;
constexpr int KMAX = 10;
constexpr int KFACMAX = 100;
constexpr int NMAX = 200010;
bool cannotCoexist[KFACMAX][KFACMAX];
int flatten(int k, int v) { return k * (k - 1) / 2 + v; };
int ans = 0, k;
int currentAns[KMAX];
bool cannotSet[KMAX][KMAX];
void computeAnswer(int i) {
  if (i > k) {
    bool valid = true;
    for (int x = 1; x <= k; ++x) {
      for (int y = 1; y < x; ++y) {
        if (cannotCoexist[flatten(x, currentAns[x])][flatten(y, currentAns[y])])
          valid = false;
      }
    }
    if (!valid) return;
    if (valid) ++ans;
  } else {
    for (int cv = 0; cv < i; ++cv) {
      if (cannotSet[i][cv]) continue;
      currentAns[i] = cv;
      computeAnswer(i + 1);
    }
  }
}
int main() {
  int n, m;
  scanf("%d %d %d", &n, &m, &k);
  vector<vector<pair<int, int>>> graph(n);
  vector<set<pair<int, int>>> satisfiers(n);
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    --u;
    --v;
    graph[u].emplace_back(w, v);
  }
  for (int i = 0; i < n; ++i) {
    sort(begin(graph[i]), end(graph[i]));
  }
  for (int i = 0; i < n; ++i) {
    int od = graph[i].size();
    for (int idx = 0; idx < od; ++idx) {
      int nxt = graph[i][idx].second;
      if (satisfiers[nxt].count(make_pair(od, idx)) > 0)
        cannotSet[od][idx] = true;
      satisfiers[nxt].emplace(od, idx);
    }
  }
  for (int i = 0; i < n; ++i) {
    for (auto [pk, pv] : satisfiers[i]) {
      for (auto [qk, qv] : satisfiers[i]) {
        cannotCoexist[flatten(pk, pv)][flatten(qk, qv)] = true;
      }
    }
  }
  computeAnswer(1);
  printf("%d\n", ans);
  return 0;
}
