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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2")
using namespace std;
using lld = long long int;
using pll = pair<lld, lld>;
using lf = double;
using pff = pair<lf, lf>;
using pii = pair<int, int>;
using pli = pair<lld, int>;
using pil = pair<int, lld>;
using piii = pair<int, pii>;
int n, m, p;
piii e[200009];
vector<int> v[200009];
vector<int> rv[10];
const lld vM[10] = {
    1000000007, 1000000009, 998244363, 1000000021, 1000000033,
};
const int ml = 5;
const int el = 11;
struct HZ {
  lld v[ml][el];
  HZ() {
    for (int i = 0; i < ml; i++) {
      for (int j = 0; j < el; j++) {
        v[i][j] = 0;
      }
    }
  }
  void add(int x) {
    for (int i = 0; i < ml; i++) {
      lld z = 1;
      for (int j = 0; j < el; j++) {
        z = (z * x) % vM[i];
        v[i][j] = (v[i][j] + z) % vM[i];
      }
    }
  }
  bool operator==(const HZ& z) {
    for (int i = 0; i < ml; i++) {
      for (int j = 0; j < el; j++) {
        if (v[i][j] != z.v[i][j]) return false;
      }
    }
    return true;
  }
  HZ operator+(const HZ& z) {
    HZ res;
    for (int i = 0; i < ml; i++) {
      for (int j = 0; j < el; j++) {
        res.v[i][j] = (v[i][j] + z.v[i][j]) % vM[i];
      }
    }
    return res;
  }
};
HZ rb[10][10];
bool rbable[10][10];
int res;
HZ az;
void dfs(int dep, HZ s) {
  if (dep == p) {
    if (az == s) {
      res++;
    }
    return;
  }
  for (int i = 0; i <= dep; i++) {
    if (rbable[dep][i]) {
      dfs(dep + 1, s + rb[dep][i]);
    }
  }
}
int main() {
  int i, j, k, l;
  int t = 1, tv = 0;
  while (t--) {
    scanf("%d %d %d", &n, &m, &p);
    for (i = 0; i < m; i++) {
      scanf("%d %d %d", &e[i].second.first, &e[i].second.second, &e[i].first);
    }
    for (i = 0; i < n; i++) {
      az.add(i);
    }
    sort(e, e + m);
    for (i = 0; i < m; i++) {
      v[e[i].second.first - 1].push_back(e[i].second.second - 1);
    }
    for (i = 0; i < n; i++) {
      rv[v[i].size()].push_back(i);
    }
    if (rv[0].size() > 0) {
      printf("0\n");
      continue;
    }
    bitset<200000> s;
    for (i = 1; i <= p; i++) {
      for (k = 0; k < i; k++) {
        bool valid = true;
        s.reset();
        for (j = 0; j < rv[i].size(); j++) {
          int ti = v[rv[i][j]][k];
          if (s[ti]) {
            valid = false;
            break;
          }
          s.set(ti);
          rb[i - 1][k].add(ti);
        }
        rbable[i - 1][k] = valid;
      }
    }
    HZ z;
    dfs(0, z);
    printf("%d\n", res);
  }
  return 0;
}
