#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int MOD = 998244353;
struct node {
  long long a, b;
  node() {}
  node(long long a, long long b) {
    this->a = a;
    this->b = b;
  }
  node operator+(const node& rhs) const {
    return node((a + rhs.a) % MOD, (b + rhs.b) % MOD);
  }
  node operator*(const node& rhs) const {
    return node(a * rhs.a % MOD, b * rhs.b % MOD);
  }
  bool operator==(const node& rhs) const { return a == rhs.a && b == rhs.b; }
};
node BASE(2333, 233333);
node h[maxn];
node ALL(0, 0);
node s[maxn][15];
vector<pair<int, int> > vec[maxn];
int n, m, k, ans;
void dfs(int u, node hh) {
  if (u > k) {
    ans += (hh == ALL ? 1 : 0);
    return;
  }
  for (int i = 0; i < u; i++) {
    node v = hh + s[u][i];
    dfs(u + 1, v);
  }
  return;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  h[0] = node(1, 1);
  for (int i = 1; i <= n; i++) {
    h[i] = h[i - 1] * BASE;
    ALL = ALL + h[i];
  }
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    vec[u].emplace_back(make_pair(v, w));
  }
  for (int i = 1; i <= n; i++) {
    sort(vec[i].begin(), vec[i].end(), [](pair<int, int> a, pair<int, int> b) {
      return a.second < b.second;
    });
    int lens = vec[i].size();
    for (int j = 0; j < lens; j++) s[lens][j] = s[lens][j] + h[vec[i][j].first];
  }
  dfs(1, node(0, 0));
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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;
const int maxn = 2e5 + 10;
const int MOD = 998244353;
struct node {
  long long a, b;
  node() {}
  node(long long a, long long b) {
    this->a = a;
    this->b = b;
  }
  node operator+(const node& rhs) const {
    return node((a + rhs.a) % MOD, (b + rhs.b) % MOD);
  }
  node operator*(const node& rhs) const {
    return node(a * rhs.a % MOD, b * rhs.b % MOD);
  }
  bool operator==(const node& rhs) const { return a == rhs.a && b == rhs.b; }
};
node BASE(2333, 233333);
node h[maxn];
node ALL(0, 0);
node s[maxn][15];
vector<pair<int, int> > vec[maxn];
int n, m, k, ans;
void dfs(int u, node hh) {
  if (u > k) {
    ans += (hh == ALL ? 1 : 0);
    return;
  }
  for (int i = 0; i < u; i++) {
    node v = hh + s[u][i];
    dfs(u + 1, v);
  }
  return;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  h[0] = node(1, 1);
  for (int i = 1; i <= n; i++) {
    h[i] = h[i - 1] * BASE;
    ALL = ALL + h[i];
  }
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    vec[u].emplace_back(make_pair(v, w));
  }
  for (int i = 1; i <= n; i++) {
    sort(vec[i].begin(), vec[i].end(), [](pair<int, int> a, pair<int, int> b) {
      return a.second < b.second;
    });
    int lens = vec[i].size();
    for (int j = 0; j < lens; j++) s[lens][j] = s[lens][j] + h[vec[i][j].first];
  }
  dfs(1, node(0, 0));
  printf("%d\n", ans);
  return 0;
}

```