#include <bits/stdc++.h>
using namespace std;
using namespace std;
mt19937 __MT(chrono::system_clock::now().time_since_epoch().count());
const int maxn = 200000 + 10;
const int hnum = 3;
const int mod = 1e9 + 7;
const int hsv[hnum] = {67, 23, 81};
inline int mul(int a, int b, int mod) {
  long long ans = 1LL * a * b;
  ans = ans - ans / mod * mod;
  if (ans < 0) ans += mod;
  return ans;
}
inline int add(int a, int b, int mod) {
  long long ans = a + b;
  ans = ans - ans / mod * mod;
  if (ans < 0) ans += mod;
  return ans;
}
vector<pair<int, int> > G[maxn];
int n, m, k;
struct node {
  int val[2][hnum];
  node(long long x = 0) { set_val(x); }
  void set_val(long long x) {
    for (auto j = (0); j < (2); ++j)
      for (auto i = (0); i < (hnum); ++i) val[j][i] = (x % mod + mod) % mod;
  }
  node operator+(const node& o) const {
    node ans = node();
    for (auto i = (0); i < (hnum); ++i) {
      ans.val[0][i] = add(val[0][i], o.val[0][i], mod);
    }
    for (auto i = (0); i < (hnum); ++i) {
      ans.val[1][i] = mul(val[1][i], o.val[1][i], mod);
    }
    return ans;
  }
  bool operator==(const node& o) const {
    for (auto i = (0); i < (2); ++i) {
      for (auto j = (0); j < (hnum); ++j) {
        if (val[i][j] != o.val[i][j]) return false;
      }
    }
    return true;
  }
};
node uset;
node val[maxn];
node c[10][10];
int dfs(int d, node now) {
  int ans = 0;
  if (d == k + 1) {
    if (now == uset) ans++;
  } else {
    for (auto i = (1); i <= (d); ++i) {
      ans += dfs(d + 1, now + c[d][i]);
    }
  }
  return ans;
}
int main(void) {
  scanf("%d%d%d", &n, &m, &k);
  for (auto i = (1); i <= (m); ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    G[u].push_back({w, v});
  }
  for (auto i = (1); i <= (n); ++i) {
    val[i] = node(__MT());
  }
  for (auto i = (1); i <= (n); ++i) uset = uset + val[i];
  for (auto i = (1); i <= (k); ++i)
    for (auto j = (1); j <= (i); ++j) c[i][j] = node();
  for (auto i = (1); i <= (n); ++i) {
    sort((G[i]).begin(), (G[i]).end());
    int x = ((int)(G[i]).size());
    for (auto j = (1); j <= (x); ++j) {
      c[x][j] = c[x][j] + val[G[i][j - 1].second];
    }
  }
  printf("%d\n", dfs(1, node()));
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
using namespace std;
mt19937 __MT(chrono::system_clock::now().time_since_epoch().count());
const int maxn = 200000 + 10;
const int hnum = 3;
const int mod = 1e9 + 7;
const int hsv[hnum] = {67, 23, 81};
inline int mul(int a, int b, int mod) {
  long long ans = 1LL * a * b;
  ans = ans - ans / mod * mod;
  if (ans < 0) ans += mod;
  return ans;
}
inline int add(int a, int b, int mod) {
  long long ans = a + b;
  ans = ans - ans / mod * mod;
  if (ans < 0) ans += mod;
  return ans;
}
vector<pair<int, int> > G[maxn];
int n, m, k;
struct node {
  int val[2][hnum];
  node(long long x = 0) { set_val(x); }
  void set_val(long long x) {
    for (auto j = (0); j < (2); ++j)
      for (auto i = (0); i < (hnum); ++i) val[j][i] = (x % mod + mod) % mod;
  }
  node operator+(const node& o) const {
    node ans = node();
    for (auto i = (0); i < (hnum); ++i) {
      ans.val[0][i] = add(val[0][i], o.val[0][i], mod);
    }
    for (auto i = (0); i < (hnum); ++i) {
      ans.val[1][i] = mul(val[1][i], o.val[1][i], mod);
    }
    return ans;
  }
  bool operator==(const node& o) const {
    for (auto i = (0); i < (2); ++i) {
      for (auto j = (0); j < (hnum); ++j) {
        if (val[i][j] != o.val[i][j]) return false;
      }
    }
    return true;
  }
};
node uset;
node val[maxn];
node c[10][10];
int dfs(int d, node now) {
  int ans = 0;
  if (d == k + 1) {
    if (now == uset) ans++;
  } else {
    for (auto i = (1); i <= (d); ++i) {
      ans += dfs(d + 1, now + c[d][i]);
    }
  }
  return ans;
}
int main(void) {
  scanf("%d%d%d", &n, &m, &k);
  for (auto i = (1); i <= (m); ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    G[u].push_back({w, v});
  }
  for (auto i = (1); i <= (n); ++i) {
    val[i] = node(__MT());
  }
  for (auto i = (1); i <= (n); ++i) uset = uset + val[i];
  for (auto i = (1); i <= (k); ++i)
    for (auto j = (1); j <= (i); ++j) c[i][j] = node();
  for (auto i = (1); i <= (n); ++i) {
    sort((G[i]).begin(), (G[i]).end());
    int x = ((int)(G[i]).size());
    for (auto j = (1); j <= (x); ++j) {
      c[x][j] = c[x][j] + val[G[i][j - 1].second];
    }
  }
  printf("%d\n", dfs(1, node()));
  return 0;
}

```