#include <bits/stdc++.h>
using namespace std;
mt19937 mt_rand(
    chrono::high_resolution_clock::now().time_since_epoch().count());
const long long hashSize = 3;
long long mod[hashSize];
const long long N = 2e5 + 50;
long long ans = 0;
struct hash_number {
  long long a[hashSize];
  hash_number() { fill(a, a + hashSize, 0ll); }
  hash_number(long long x) {
    for (long long i = (0); i <= (hashSize - 1); ++i) {
      a[i] = x % mod[i];
    }
  }
  hash_number operator+(hash_number x) {
    hash_number res = hash_number();
    for (long long i = (0); i <= (hashSize - 1); ++i) {
      res.a[i] = (a[i] + x.a[i]) % mod[i];
    }
    return res;
  }
  bool operator==(const hash_number &x) {
    for (long long i = (0); i <= (hashSize - 1); ++i) {
      if (a[i] != x.a[i]) return false;
    }
    return true;
  }
} val[N], c[10][10], s;
vector<pair<long long, long long> > g[N];
long long n, m, k;
void solve(long long lvl, hash_number hashNumber) {
  if (lvl == k) {
    if (hashNumber == s) ++ans;
    return;
  }
  for (long long i = 1; i <= lvl + 1; i++) {
    solve(lvl + 1, hashNumber + c[lvl + 1][i]);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  mod[0] = 1e9 + 7;
  mod[1] = 998244353;
  mod[2] = uniform_int_distribution<long long>(1e8, 1e9)(mt_rand);
  cin >> n >> m >> k;
  for (long long i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    g[u].emplace_back(w, v);
  }
  uniform_int_distribution<long long> rg(1ll, 1e18);
  for (long long i = 1; i <= n; i++) {
    val[i] = hash_number(rg(mt_rand));
    s = s + val[i];
  }
  for (long long u = 1; u <= n; u++) {
    long long d = g[u].size();
    sort(g[u].begin(), g[u].end());
    for (long long q = 1; q <= d; q++) {
      c[d][q] = c[d][q] + val[g[u][q - 1].second];
    }
  }
  solve(0, hash_number());
  cout << ans << '\n';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937 mt_rand(
    chrono::high_resolution_clock::now().time_since_epoch().count());
const long long hashSize = 3;
long long mod[hashSize];
const long long N = 2e5 + 50;
long long ans = 0;
struct hash_number {
  long long a[hashSize];
  hash_number() { fill(a, a + hashSize, 0ll); }
  hash_number(long long x) {
    for (long long i = (0); i <= (hashSize - 1); ++i) {
      a[i] = x % mod[i];
    }
  }
  hash_number operator+(hash_number x) {
    hash_number res = hash_number();
    for (long long i = (0); i <= (hashSize - 1); ++i) {
      res.a[i] = (a[i] + x.a[i]) % mod[i];
    }
    return res;
  }
  bool operator==(const hash_number &x) {
    for (long long i = (0); i <= (hashSize - 1); ++i) {
      if (a[i] != x.a[i]) return false;
    }
    return true;
  }
} val[N], c[10][10], s;
vector<pair<long long, long long> > g[N];
long long n, m, k;
void solve(long long lvl, hash_number hashNumber) {
  if (lvl == k) {
    if (hashNumber == s) ++ans;
    return;
  }
  for (long long i = 1; i <= lvl + 1; i++) {
    solve(lvl + 1, hashNumber + c[lvl + 1][i]);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  mod[0] = 1e9 + 7;
  mod[1] = 998244353;
  mod[2] = uniform_int_distribution<long long>(1e8, 1e9)(mt_rand);
  cin >> n >> m >> k;
  for (long long i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    g[u].emplace_back(w, v);
  }
  uniform_int_distribution<long long> rg(1ll, 1e18);
  for (long long i = 1; i <= n; i++) {
    val[i] = hash_number(rg(mt_rand));
    s = s + val[i];
  }
  for (long long u = 1; u <= n; u++) {
    long long d = g[u].size();
    sort(g[u].begin(), g[u].end());
    for (long long q = 1; q <= d; q++) {
      c[d][q] = c[d][q] + val[g[u][q - 1].second];
    }
  }
  solve(0, hash_number());
  cout << ans << '\n';
}

```