#include <bits/stdc++.h>
using namespace std;
vector<long long> p1, p2;
vector<vector<pair<int, int>>> v;
vector<vector<int>> degree;
int A1 = 100007, B1 = 999999997;
int A2 = 54839, B2 = 1000000007;
int ans = 0, ans1 = 0, ans2 = 0;
int N, M, K;
vector<vector<int>> hashset1, hashset2;
void myfunc(int icount, int n1, int n2) {
  if (icount > K) {
    if (n1 == ans1 && n2 == ans2) ans++;
    return;
  }
  for (int i = 0; i < icount; ++i) {
    myfunc(icount + 1, (n1 + hashset1[icount][i]) % B1,
           (n2 + hashset2[icount][i]) % B2);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M >> K;
  v = vector<vector<pair<int, int>>>(N + 1);
  for (int i = 0; i < M; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    v[a].push_back({c, b});
  }
  for (auto& x : v) sort(x.begin(), x.end());
  degree = vector<vector<int>>(K + 1);
  for (int n = 1; n <= N; ++n) {
    degree[v[n].size()].push_back(n);
  }
  p1 = vector<long long>(N + 1);
  p2 = vector<long long>(N + 1);
  p1[1] = A1;
  p2[1] = A2;
  for (int n = 2; n <= N; ++n) {
    p1[n] = (p1[n - 1] * A1) % B1;
    p2[n] = (p2[n - 1] * A2) % B2;
  }
  for (int n = 1; n <= N; ++n) {
    ans1 = (ans1 + p1[n]) % B1;
    ans2 = (ans2 + p2[n]) % B2;
  }
  hashset1 = vector<vector<int>>(K + 1, vector<int>(K + 1));
  hashset2 = vector<vector<int>>(K + 1, vector<int>(K + 1));
  for (int i = 1; i <= K; ++i) {
    for (int j = 0; j < i; ++j) {
      for (auto x : degree[i]) {
        hashset1[i][j] = (hashset1[i][j] + p1[v[x][j].second]) % B1;
        hashset2[i][j] = (hashset2[i][j] + p2[v[x][j].second]) % B2;
      }
    }
  }
  myfunc(1, 0, 0);
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
vector<long long> p1, p2;
vector<vector<pair<int, int>>> v;
vector<vector<int>> degree;
int A1 = 100007, B1 = 999999997;
int A2 = 54839, B2 = 1000000007;
int ans = 0, ans1 = 0, ans2 = 0;
int N, M, K;
vector<vector<int>> hashset1, hashset2;
void myfunc(int icount, int n1, int n2) {
  if (icount > K) {
    if (n1 == ans1 && n2 == ans2) ans++;
    return;
  }
  for (int i = 0; i < icount; ++i) {
    myfunc(icount + 1, (n1 + hashset1[icount][i]) % B1,
           (n2 + hashset2[icount][i]) % B2);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M >> K;
  v = vector<vector<pair<int, int>>>(N + 1);
  for (int i = 0; i < M; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    v[a].push_back({c, b});
  }
  for (auto& x : v) sort(x.begin(), x.end());
  degree = vector<vector<int>>(K + 1);
  for (int n = 1; n <= N; ++n) {
    degree[v[n].size()].push_back(n);
  }
  p1 = vector<long long>(N + 1);
  p2 = vector<long long>(N + 1);
  p1[1] = A1;
  p2[1] = A2;
  for (int n = 2; n <= N; ++n) {
    p1[n] = (p1[n - 1] * A1) % B1;
    p2[n] = (p2[n - 1] * A2) % B2;
  }
  for (int n = 1; n <= N; ++n) {
    ans1 = (ans1 + p1[n]) % B1;
    ans2 = (ans2 + p2[n]) % B2;
  }
  hashset1 = vector<vector<int>>(K + 1, vector<int>(K + 1));
  hashset2 = vector<vector<int>>(K + 1, vector<int>(K + 1));
  for (int i = 1; i <= K; ++i) {
    for (int j = 0; j < i; ++j) {
      for (auto x : degree[i]) {
        hashset1[i][j] = (hashset1[i][j] + p1[v[x][j].second]) % B1;
        hashset2[i][j] = (hashset2[i][j] + p2[v[x][j].second]) % B2;
      }
    }
  }
  myfunc(1, 0, 0);
  cout << ans;
  return 0;
}

```