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 int mod1 = 1e9 + 7;
const int mod2 = 998244353;
long long c[100][100], c2[100][100];
vector<pair<int, int> > g[200010];
bool prime[5000000];
int p[500000], tot = 0;
void gt_prime() {
  for (int i = 2; i < 5000000; i++) {
    if (!prime[i]) p[++tot] = i;
    for (int j = 1; j <= tot && p[j] * i < 5000000; j++) {
      prime[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
int main() {
  gt_prime();
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int u, v, k;
    scanf("%d%d%d", &u, &v, &k);
    g[u].push_back({k, v});
  }
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      c[i][j] = 1;
      c2[i][j] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    int nl = (int)g[i].size();
    if (nl > 1) sort(g[i].begin(), g[i].end());
    for (int j = 0; j < nl; j++) {
      int x = g[i][j].second;
      c[nl][j + 1] *= p[x];
      c[nl][j + 1] %= mod1;
      c2[nl][j + 1] *= p[x];
      c2[nl][j + 1] %= mod2;
    }
  }
  int d[19];
  for (int i = 1; i <= 9; i++) {
    d[i] = 1;
  }
  long long res1 = 1;
  long long res2 = 1;
  for (int i = 1; i <= n; i++) {
    res1 = (res1 * p[i]) % mod1;
    res2 = (res2 * p[i]) % mod2;
  }
  int cnt = 0, cishu = 0;
  while (1) {
    long long ans1 = 1;
    long long ans2 = 1;
    for (int i = 1; i <= 9; i++) {
      ans1 = ans1 * c[i][d[i]] % mod1;
      ans2 = ans2 * c2[i][d[i]] % mod2;
    }
    if (ans1 == res1 && ans2 == res2) cnt++;
    cishu++;
    if (cishu == 362880) break;
    d[9]++;
    int pxr = 9;
    while (pxr != 1 && d[pxr] > pxr) {
      d[pxr] = 1;
      d[pxr - 1]++;
      pxr--;
    }
    if (d[1] == 2) break;
  }
  for (int i = k + 1; i <= 9; i++) cnt /= i;
  cout << cnt << endl;
  return 0;
}
