#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool readn(T &x);
void rd();
void debug();
const int N = 5e5 + 9;
const long long mod = 1e9 + 7;
int T, n, m, k;
const int M = 15;
unsigned long long tot, hs[N], mp[M][M];
struct E {
  int to;
  int w;
};
vector<E> v[N];
vector<int> e[M];
bool cmp(E A, E B) { return A.w < B.w; }
int ans = 0;
void dfs(int nowp, unsigned long long res) {
  if (nowp == k + 1) {
    if (res == tot) {
      ++ans;
    }
    return;
  }
  for (int i = 1; i <= nowp; ++i) {
    dfs(nowp + 1, res + mp[nowp][i]);
  }
}
void solve() {
  rd();
  for (T = 1; T; T--) {
    readn(n);
    readn(m);
    readn(k);
    hs[1] = 1;
    for (int i = 2; i <= n; ++i) {
      hs[i] = hs[i - 1] * 2;
    }
    for (int i = 1; i <= n; ++i) tot = tot + hs[i];
    for (int i = 1; i <= m; ++i) {
      int x, y, w;
      readn(x);
      readn(y);
      readn(w);
      v[x].push_back({y, w});
    }
    for (int i = 1; i <= n; ++i) {
      sort(v[i].begin(), v[i].end(), cmp);
      int sz = v[i].size();
      e[sz].push_back(i);
    }
    for (int i = 1; i <= k; ++i) {
      for (int j = 1; j <= i; ++j) {
        for (int p = 0; p < e[i].size(); ++p) {
          int to = e[i][p];
          mp[i][j] = mp[i][j] + hs[v[to][j - 1].to];
        }
      }
    }
    dfs(1, 0);
    printf("%d\n", ans);
  }
  return;
}
int main() {
  solve();
  return 0;
}
template <typename T>
inline bool readn(T &x) {
  bool op = 0;
  char ch = getchar();
  x = 0;
  for (; !isdigit(ch) && ch != EOF; ch = getchar()) op ^= (ch == '-');
  if (ch == EOF) return false;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ '0');
  if (op) x = -x;
  return true;
}
void rd() {}
void debug() { puts("*******************************"); }

### Prompt

Your challenge is to write 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;
template <typename T>
inline bool readn(T &x);
void rd();
void debug();
const int N = 5e5 + 9;
const long long mod = 1e9 + 7;
int T, n, m, k;
const int M = 15;
unsigned long long tot, hs[N], mp[M][M];
struct E {
  int to;
  int w;
};
vector<E> v[N];
vector<int> e[M];
bool cmp(E A, E B) { return A.w < B.w; }
int ans = 0;
void dfs(int nowp, unsigned long long res) {
  if (nowp == k + 1) {
    if (res == tot) {
      ++ans;
    }
    return;
  }
  for (int i = 1; i <= nowp; ++i) {
    dfs(nowp + 1, res + mp[nowp][i]);
  }
}
void solve() {
  rd();
  for (T = 1; T; T--) {
    readn(n);
    readn(m);
    readn(k);
    hs[1] = 1;
    for (int i = 2; i <= n; ++i) {
      hs[i] = hs[i - 1] * 2;
    }
    for (int i = 1; i <= n; ++i) tot = tot + hs[i];
    for (int i = 1; i <= m; ++i) {
      int x, y, w;
      readn(x);
      readn(y);
      readn(w);
      v[x].push_back({y, w});
    }
    for (int i = 1; i <= n; ++i) {
      sort(v[i].begin(), v[i].end(), cmp);
      int sz = v[i].size();
      e[sz].push_back(i);
    }
    for (int i = 1; i <= k; ++i) {
      for (int j = 1; j <= i; ++j) {
        for (int p = 0; p < e[i].size(); ++p) {
          int to = e[i][p];
          mp[i][j] = mp[i][j] + hs[v[to][j - 1].to];
        }
      }
    }
    dfs(1, 0);
    printf("%d\n", ans);
  }
  return;
}
int main() {
  solve();
  return 0;
}
template <typename T>
inline bool readn(T &x) {
  bool op = 0;
  char ch = getchar();
  x = 0;
  for (; !isdigit(ch) && ch != EOF; ch = getchar()) op ^= (ch == '-');
  if (ch == EOF) return false;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ '0');
  if (op) x = -x;
  return true;
}
void rd() {}
void debug() { puts("*******************************"); }

```