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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gen() { return 2 * (500000 * (rng() % 5 + 1) + rng() % 5000000) + 1; }
long long base;
long long binpow(long long x, long long y) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret *= x;
    x *= x;
    x %= 1000000007;
    ret %= 1000000007;
    y /= 2;
  }
  return ret;
}
long long baash(long long x) { return (binpow(base, x) * x) % 1000000007; }
long long arr[15][15];
vector<pair<long long, long long> > adj[200005];
long long vis[200005], org;
void dfs(long long n) {
  vis[n] = 1;
  for (long long i = 0; i < adj[n].size(); i++) {
    auto &cv = adj[n][i];
    arr[adj[n].size()][i + 1] += baash(cv.second);
    arr[adj[n].size()][i + 1] %= 1000000007;
    if (vis[cv.second]) continue;
    dfs(cv.second);
  }
}
long long ans;
void bct(long long i, long long cur) {
  cur %= 1000000007;
  if (i == 0) {
    if (cur == org) ans++;
    return;
  }
  for (long long j = 1; j <= i; j++) {
    bct(i - 1, cur + arr[i][j]);
  }
}
void solve() {
  long long i, j, k, l, r, m, n, a, b, c, d, x, y, z, t;
  base = gen();
  cin >> n >> m >> k;
  for (i = 0; i < m; i++) {
    cin >> x >> y >> z;
    adj[x].push_back({z, y});
  }
  for (i = 1; i <= n; i++) {
    sort(adj[i].begin(), adj[i].end());
  }
  for (i = 1; i <= n; i++) {
    if (vis[i]) continue;
    dfs(i);
  }
  for (i = 1; i <= n; i++) {
    org += baash(i);
    org %= 1000000007;
  }
  x = arr[1][1];
  bct(k, 0);
  cout << ans << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}
