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 MOD = (long long)1e9 + 7;
const long double PI = 3.141592653589793238462643383279502884197;
long long fac[1] = {1}, inv[1] = {1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long mp(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % MOD;
    a = a * a % MOD;
    b >>= 1;
  }
  return ret;
}
long long cmb(long long r, long long c) {
  return fac[r] * inv[c] % MOD * inv[r - c] % MOD;
}
int n, m, k;
int cant[10][10];
int mat[10][10][10][10];
int visit[10][10];
vector<pair<int, int> > vv[200000];
vector<pair<int, int> > v[200000];
int od[200000];
int ans = 0;
int now[10];
void go(int o) {
  if (o == k + 1) {
    ans++;
    return;
  }
  for (int j = 1; j <= o; j++) {
    if (cant[o][j]) continue;
    int flag = 0;
    for (int kk = 1; kk < o; kk++)
      if (mat[kk][now[kk]][o][j]) {
        flag = 1;
        break;
      }
    if (flag == 1) continue;
    now[o] = j;
    go(o + 1);
    now[o] = 0;
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    a--;
    b--;
    od[a]++;
    vv[a].push_back({c, b});
  }
  for (int i = 0; i < n; i++) {
    sort((vv[i]).begin(), (vv[i]).end());
    for (int j = 0; j < (int)(vv[i]).size(); j++)
      v[vv[i][j].second].push_back({od[i], j + 1});
  }
  for (int i = 0; i < n; i++) {
    memset(visit, 0, sizeof(visit));
    for (pair<int, int> j : v[i]) {
      visit[j.first][j.second]++;
      if (visit[j.first][j.second] >= 2) cant[j.first][j.second] = 1;
    }
    if ((int)(v[i]).size() >= 101) {
      for (int a = 1; a <= k; a++)
        for (int b = 1; b <= a; b++)
          for (int c = 1; c <= a; c++)
            for (int d = 1; d <= c; d++)
              if (visit[a][b] && visit[c][d])
                mat[c][d][a][b] = mat[a][b][c][d] = 1;
    } else {
      for (int j = 0; j < (int)(v[i]).size(); j++)
        for (int kk = 0; kk < j; kk++) {
          mat[v[i][j].first][v[i][j].second][v[i][kk].first][v[i][kk].second] =
              1;
          mat[v[i][kk].first][v[i][kk].second][v[i][j].first][v[i][j].second] =
              1;
        }
    }
  }
  go(1);
  printf("%d", ans);
}
