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>
const int maxn = 2e5 + 5;
const int p = 131;
const int mo = (1 << 21) - 1;
const int maxk = 10;
using namespace std;
int rd() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return x * f;
}
int n, m, k, c[maxn], ans, mp[maxk][maxk], e;
vector<pair<int, int> > t[maxn];
void add(int &x, int y) {
  x += y;
  if (x >= mo) x -= mo;
}
bool tmp(pair<int, int> &x, pair<int, int> &y) { return x.second > y.second; }
bool check() {
  int now = 0;
  for (int i = 1; i <= k; i++) add(now, mp[i][c[i]]);
  return now == e;
}
void dfs(int x) {
  if (x == k + 1) {
    if (check()) ans++;
    return;
  }
  for (int i = 1; i <= x; i++) {
    c[x] = i;
    dfs(x + 1);
  }
}
int ksm(int a, int b) {
  int sum = 1;
  for (; b; b >>= 1, a = 1LL * a * a % mo)
    if (b & 1) sum = 1LL * sum * a % mo;
  return sum;
}
int main() {
  n = rd();
  m = rd();
  k = rd();
  for (int u, v, w, i = 1; i <= m; i++) {
    u = rd();
    v = rd();
    w = rd();
    t[u].push_back(make_pair(v, w));
  }
  for (int i = 1; i <= n; i++) {
    sort(t[i].begin(), t[i].end(), tmp);
    int s = t[i].size();
    for (int j = 0; j < s; j++) add(mp[s][j + 1], ksm(p, t[i][j].first));
  }
  for (int i = 1; i <= n; i++) add(e, ksm(p, i));
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
