#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int n, m, k;
struct node {
  int to, nxt, key;
} edge[maxn << 1];
int head[maxn], tot = 0;
int out[maxn];
int hash1[maxn], hash2[maxn];
int hash3[maxn], hash4[maxn];
vector<int> vv[15];
struct node1 {
  int v, c;
  bool operator<(const node1 a) const { return c > a.c; }
};
set<node1> sset[maxn];
set<node1>::iterator it;
long long ans1[15][15];
long long ans2[15][15];
long long ans3[15][15];
long long ans4[15][15];
inline void add(int u, int v, int w) { sset[u].insert((node1){v, w}); }
long long num1, num2, num3, num4;
inline void init() {
  srand(time(0));
  n = read();
  m = read();
  k = read();
  num1 = num2 = num3 = num4 = 1;
  for (int i = 1; i <= n; ++i)
    hash1[i] = rand(), hash2[i] = rand(), num1 = num1 * hash1[i] % mod,
    num2 = num2 * hash2[i] % mod;
  for (int i = 1; i <= n; ++i)
    hash3[i] = rand(), hash4[i] = rand(), num3 = num3 * hash3[i] % mod,
    num4 = num4 * hash4[i] % mod;
  for (int i = 1; i <= n; ++i) sset[i].clear(), out[i] = 0;
  for (int i = 1; i <= m; ++i) {
    int u = read();
    int v = read();
    int w = read();
    add(u, v, w);
    ++out[u];
  }
  for (int i = 1; i <= k; ++i) vv[i].clear();
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= k; ++j)
      ans1[i][j] = ans2[i][j] = ans3[i][j] = ans4[i][j] = 1;
}
int st[maxn];
int num = 0;
long long res;
inline void dfs(int step) {
  if (step == k) {
    long long tmp = 1;
    int flag = 0;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans1[i][j] % mod;
    }
    if (tmp != num1) flag = 1;
    tmp = 1;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans2[i][j] % mod;
    }
    if (tmp != num2) flag = 1;
    tmp = 1;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans3[i][j] % mod;
    }
    if (tmp != num3) flag = 1;
    tmp = 1;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans4[i][j] % mod;
    }
    if (tmp != num4) flag = 1;
    if (!flag) ++res;
    return;
  }
  for (int i = 1; i <= step + 1; ++i) {
    st[++num] = i;
    dfs(step + 1);
    --num;
  }
}
inline void solve() {
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= n; ++j)
      if (out[j] == i) vv[i].push_back(j);
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < vv[i].size(); ++j) {
      int tmp = vv[i][j];
      int o = 0;
      for (it = sset[tmp].begin(); it != sset[tmp].end(); ++it) {
        ++o;
        int v = (*it).v;
        ans1[i][o] = ans1[i][o] * hash1[v] % mod;
        ans2[i][o] = ans2[i][o] * hash2[v] % mod;
        ans3[i][o] = ans3[i][o] * hash3[v] % mod;
        ans4[i][o] = ans4[i][o] * hash4[v] % mod;
      }
    }
  }
  res = 0;
  dfs(0);
  printf("%lld\n", res);
}
int main() {
  init();
  solve();
}

### Prompt

Please create a solution in cpp 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(2)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
int n, m, k;
struct node {
  int to, nxt, key;
} edge[maxn << 1];
int head[maxn], tot = 0;
int out[maxn];
int hash1[maxn], hash2[maxn];
int hash3[maxn], hash4[maxn];
vector<int> vv[15];
struct node1 {
  int v, c;
  bool operator<(const node1 a) const { return c > a.c; }
};
set<node1> sset[maxn];
set<node1>::iterator it;
long long ans1[15][15];
long long ans2[15][15];
long long ans3[15][15];
long long ans4[15][15];
inline void add(int u, int v, int w) { sset[u].insert((node1){v, w}); }
long long num1, num2, num3, num4;
inline void init() {
  srand(time(0));
  n = read();
  m = read();
  k = read();
  num1 = num2 = num3 = num4 = 1;
  for (int i = 1; i <= n; ++i)
    hash1[i] = rand(), hash2[i] = rand(), num1 = num1 * hash1[i] % mod,
    num2 = num2 * hash2[i] % mod;
  for (int i = 1; i <= n; ++i)
    hash3[i] = rand(), hash4[i] = rand(), num3 = num3 * hash3[i] % mod,
    num4 = num4 * hash4[i] % mod;
  for (int i = 1; i <= n; ++i) sset[i].clear(), out[i] = 0;
  for (int i = 1; i <= m; ++i) {
    int u = read();
    int v = read();
    int w = read();
    add(u, v, w);
    ++out[u];
  }
  for (int i = 1; i <= k; ++i) vv[i].clear();
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= k; ++j)
      ans1[i][j] = ans2[i][j] = ans3[i][j] = ans4[i][j] = 1;
}
int st[maxn];
int num = 0;
long long res;
inline void dfs(int step) {
  if (step == k) {
    long long tmp = 1;
    int flag = 0;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans1[i][j] % mod;
    }
    if (tmp != num1) flag = 1;
    tmp = 1;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans2[i][j] % mod;
    }
    if (tmp != num2) flag = 1;
    tmp = 1;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans3[i][j] % mod;
    }
    if (tmp != num3) flag = 1;
    tmp = 1;
    for (int i = 1; i <= k; ++i) {
      int j = st[i];
      tmp = tmp * ans4[i][j] % mod;
    }
    if (tmp != num4) flag = 1;
    if (!flag) ++res;
    return;
  }
  for (int i = 1; i <= step + 1; ++i) {
    st[++num] = i;
    dfs(step + 1);
    --num;
  }
}
inline void solve() {
  for (int i = 1; i <= k; ++i) {
    for (int j = 1; j <= n; ++j)
      if (out[j] == i) vv[i].push_back(j);
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < vv[i].size(); ++j) {
      int tmp = vv[i][j];
      int o = 0;
      for (it = sset[tmp].begin(); it != sset[tmp].end(); ++it) {
        ++o;
        int v = (*it).v;
        ans1[i][o] = ans1[i][o] * hash1[v] % mod;
        ans2[i][o] = ans2[i][o] * hash2[v] % mod;
        ans3[i][o] = ans3[i][o] * hash3[v] % mod;
        ans4[i][o] = ans4[i][o] * hash4[v] % mod;
      }
    }
  }
  res = 0;
  dfs(0);
  printf("%lld\n", res);
}
int main() {
  init();
  solve();
}

```