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.
#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long __64inf = 0x3f3f3f3f3f3f3f3f;
const int MAX = 2e6 + 500;
const int mod = 1e9 + 7;
void file_read() {}
const int mod2 = 998244353;
vector<pair<int, int> > G[MAX];
unsigned long long s[100][100];
long long g[100][100];
long long t[100][100];
bool vis[MAX];
unsigned long long val[MAX];
unsigned long long TARGET1 = 1;
long long TARGET2 = 1;
long long TARGET3 = 1;
int n, m, k;
int ans;
void dfs(int u) {
  vis[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i].first;
    s[G[u].size()][i + 1] *= (unsigned long long)val[v];
    g[G[u].size()][i + 1] = g[G[u].size()][i + 1] * val[v] % mod;
    t[G[u].size()][i + 1] = t[G[u].size()][i + 1] * val[v] % mod2;
    if (vis[v]) continue;
    dfs(v);
  }
}
void dfs2(int x, unsigned long long cur1, long long cur2, long long cur3) {
  if (x > k) {
    if (cur1 == TARGET1 and cur2 == TARGET2 and cur3 == TARGET3) ans++;
    return;
  }
  for (int i = 1; i <= x; i++) {
    dfs2(x + 1, cur1 * s[x][i], cur2 * g[x][i] % mod, cur3 * t[x][i] % mod2);
  }
}
int main() {
  srand((int)time(0));
  file_read();
  for (int i = 0; i < 100; i++)
    for (int j = 0; j < 100; j++) s[i][j] = g[i][j] = t[i][j] = 1;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= n; i++) val[i] = rand() % mod;
  for (unsigned long long i = 1; i <= n; i++)
    TARGET1 = TARGET1 * val[i], TARGET2 = TARGET2 * val[i] % mod,
    TARGET3 = TARGET3 * val[i] % mod2;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    G[u].push_back({v, w});
  }
  for (int i = 1; i <= n; i++) {
    sort(G[i].begin(), G[i].end(),
         [](const pair<int, int>& A, const pair<int, int>& B) {
           return A.second < B.second;
         });
  }
  dfs(1);
  dfs2(1, 1, 1, 1);
  printf("%d\n", ans);
  return 0;
}
