#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
using namespace std;
int n, m, k, ct[200010], cnt;
vector<pair<int, int> > e[200010];
vector<int> mp[200010];
char bf[100000], *p1 = bf, *p2 = bf;
inline int read() {
  int x = 0;
  char ch =
      (p1 == p2 && (p2 = (p1 = bf) + fread(bf, 1, 100000, stdin), p2 == p1)
           ? -1
           : *p1++);
  for (; ch < '0' || ch > '9';
       ch =
           (p1 == p2 && (p2 = (p1 = bf) + fread(bf, 1, 100000, stdin), p2 == p1)
                ? -1
                : *p1++))
    ;
  for (; ch <= '9' && ch >= '0';
       x = x * 10 + ch - 48,
       ch =
           (p1 == p2 && (p2 = (p1 = bf) + fread(bf, 1, 100000, stdin), p2 == p1)
                ? -1
                : *p1++))
    ;
  return x;
}
int ans, O = 1;
int hah, val[10][200010];
void dfs(int x, int f) {
  if (x == k + 1) {
    if (f == hah) ++ans;
    return;
  }
  if (mp[x].size() == 0) {
    dfs(x + 1, f);
    return;
  }
  for (auto i = (1); i <= (x); ++i) {
    int flag = 1;
    dfs(x + 1, f + val[x][i]);
  }
}
int hsh[200010];
inline void solve() {
  n = read(), m = read(), k = read();
  for (auto i = (1); i <= (n); ++i) hsh[i] = rand(), hah += hsh[i];
  for (auto i = (1); i <= (m); ++i) {
    int u = read(), v = read(), w = read();
    e[u].push_back({w, v});
  }
  for (auto i = (1); i <= (n); ++i) sort(e[i].begin(), e[i].end());
  for (auto i = (1); i <= (n); ++i) mp[e[i].size()].push_back(i);
  for (auto i = (1); i <= (k); ++i)
    for (auto j = (1); j <= (i); ++j) {
      val[i][j] = 0;
      for (auto v : mp[i]) val[i][j] += hsh[e[v][j - 1].second];
    }
  for (auto i = (1); i <= (k); ++i)
    if (mp[i].size() == 0) O *= i;
  dfs(1, 0);
  printf("%d\n", ans * O);
}
int main() {
  int T = 1;
  while (T--) solve();
}

### Prompt

Generate 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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
using namespace std;
int n, m, k, ct[200010], cnt;
vector<pair<int, int> > e[200010];
vector<int> mp[200010];
char bf[100000], *p1 = bf, *p2 = bf;
inline int read() {
  int x = 0;
  char ch =
      (p1 == p2 && (p2 = (p1 = bf) + fread(bf, 1, 100000, stdin), p2 == p1)
           ? -1
           : *p1++);
  for (; ch < '0' || ch > '9';
       ch =
           (p1 == p2 && (p2 = (p1 = bf) + fread(bf, 1, 100000, stdin), p2 == p1)
                ? -1
                : *p1++))
    ;
  for (; ch <= '9' && ch >= '0';
       x = x * 10 + ch - 48,
       ch =
           (p1 == p2 && (p2 = (p1 = bf) + fread(bf, 1, 100000, stdin), p2 == p1)
                ? -1
                : *p1++))
    ;
  return x;
}
int ans, O = 1;
int hah, val[10][200010];
void dfs(int x, int f) {
  if (x == k + 1) {
    if (f == hah) ++ans;
    return;
  }
  if (mp[x].size() == 0) {
    dfs(x + 1, f);
    return;
  }
  for (auto i = (1); i <= (x); ++i) {
    int flag = 1;
    dfs(x + 1, f + val[x][i]);
  }
}
int hsh[200010];
inline void solve() {
  n = read(), m = read(), k = read();
  for (auto i = (1); i <= (n); ++i) hsh[i] = rand(), hah += hsh[i];
  for (auto i = (1); i <= (m); ++i) {
    int u = read(), v = read(), w = read();
    e[u].push_back({w, v});
  }
  for (auto i = (1); i <= (n); ++i) sort(e[i].begin(), e[i].end());
  for (auto i = (1); i <= (n); ++i) mp[e[i].size()].push_back(i);
  for (auto i = (1); i <= (k); ++i)
    for (auto j = (1); j <= (i); ++j) {
      val[i][j] = 0;
      for (auto v : mp[i]) val[i][j] += hsh[e[v][j - 1].second];
    }
  for (auto i = (1); i <= (k); ++i)
    if (mp[i].size() == 0) O *= i;
  dfs(1, 0);
  printf("%d\n", ans * O);
}
int main() {
  int T = 1;
  while (T--) solve();
}

```