#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const double eps = 1e-8;
const double pi = acos(-1);
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, d, m, k, ans;
int du[200005];
struct nd {
  int v, w;
};
vector<nd> e[200005];
vector<int> dd[10];
bool cmp(nd a, nd b) { return a.w < b.w; }
unsigned long long p[15][15], num[200005], bas = 19491001, sum, now;
void dfs(int ce) {
  int i, j, v;
  bool flag = 0;
  if (ce == k + 1) {
    ans += (now == sum * 2);
    return;
  }
  for (i = 1; i <= ce; i++) {
    now += p[i][ce];
    dfs(ce + 1);
    now -= p[i][ce];
  }
}
int main() {
  int i, j, u, v, w;
  n = read(), m = read(), k = read();
  num[0] = 1;
  for (i = 1; i <= n; i++) num[i] = num[i - 1] * bas, sum += num[i];
  for (i = 1; i <= m; i++) {
    u = read(), v = read(), w = read();
    du[u]++;
    e[u].push_back((nd){v, w});
  }
  for (i = 1; i <= n; i++) sort(e[i].begin(), e[i].end(), cmp);
  for (i = 1; i <= n; i++) dd[du[i]].push_back(i);
  for (i = 1; i <= k; i++)
    for (j = 1; j <= i; j++) {
      for (w = 0; w < dd[i].size(); w++) {
        u = dd[i][w];
        v = e[u][j - 1].v;
        p[j][i] += num[u] + num[v];
      }
    }
  dfs(1);
  cout << ans;
  return 0;
}

### 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;
const long long mod = 998244353;
const double eps = 1e-8;
const double pi = acos(-1);
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, d, m, k, ans;
int du[200005];
struct nd {
  int v, w;
};
vector<nd> e[200005];
vector<int> dd[10];
bool cmp(nd a, nd b) { return a.w < b.w; }
unsigned long long p[15][15], num[200005], bas = 19491001, sum, now;
void dfs(int ce) {
  int i, j, v;
  bool flag = 0;
  if (ce == k + 1) {
    ans += (now == sum * 2);
    return;
  }
  for (i = 1; i <= ce; i++) {
    now += p[i][ce];
    dfs(ce + 1);
    now -= p[i][ce];
  }
}
int main() {
  int i, j, u, v, w;
  n = read(), m = read(), k = read();
  num[0] = 1;
  for (i = 1; i <= n; i++) num[i] = num[i - 1] * bas, sum += num[i];
  for (i = 1; i <= m; i++) {
    u = read(), v = read(), w = read();
    du[u]++;
    e[u].push_back((nd){v, w});
  }
  for (i = 1; i <= n; i++) sort(e[i].begin(), e[i].end(), cmp);
  for (i = 1; i <= n; i++) dd[du[i]].push_back(i);
  for (i = 1; i <= k; i++)
    for (j = 1; j <= i; j++) {
      for (w = 0; w < dd[i].size(); w++) {
        u = dd[i][w];
        v = e[u][j - 1].v;
        p[j][i] += num[u] + num[v];
      }
    }
  dfs(1);
  cout << ans;
  return 0;
}

```