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;
int N, M, K;
vector<pair<int, int>> e[200005];
bool comp(pair<int, int>& a, pair<int, int>& b) { return a.second < b.second; }
struct way {
  int f, t;
};
vector<way> bin[10][10];
pair<long long, long long> has[200005];
pair<long long, long long> hbin[10][10];
pair<long long, long long> sum;
pair<long long, long long> sub;
void add(int x, int y) {
  sub.first = (sub.first + hbin[x][y].first) % 998244353;
  sub.second = (sub.second + hbin[x][y].second) % 1000000007;
}
void del(int x, int y) {
  sub.first = (sub.first - hbin[x][y].first + 998244353) % 998244353;
  sub.second = (sub.second - hbin[x][y].second + 1000000007) % 1000000007;
}
int ans = 0;
void dfs(int dep) {
  if (dep > K) {
    if (sub == sum) ++ans;
    return;
  }
  for (int i = 1; i <= dep; i++) {
    add(dep, i);
    dfs(dep + 1);
    del(dep, i);
  }
}
int main() {
  scanf("%d %d %d", &N, &M, &K);
  has[0] = {691787, 691787};
  sum = {0, 0};
  for (int i = 1; i <= N; i++) {
    has[i].first = has[i - 1].first * 19260817 % 998244353;
    has[i].second = has[i - 1].second * 19260817 % 1000000007;
    sum = {(sum.first + has[i].first) % 998244353,
           (sum.second + has[i].second) % 1000000007};
  }
  for (int i = 1; i <= M; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    e[a].push_back({b, c});
  }
  for (int i = 1; i <= N; i++) {
    sort(e[i].begin(), e[i].end(), comp);
    int di = e[i].size();
    for (int j = 0; j < di; j++) {
      bin[di][j + 1].push_back((way){i, e[i][j].first});
    }
  }
  for (int i = 1; i <= K; i++)
    for (int j = 1; j <= i; j++) {
      long long b1 = 0, b2 = 0;
      for (int k = 0; k < bin[i][j].size(); k++) {
        b1 = (b1 + has[bin[i][j][k].t].first) % 998244353;
        b2 = (b2 + has[bin[i][j][k].t].second) % 1000000007;
      }
      hbin[i][j] = {b1, b2};
    }
  sub = {0, 0};
  dfs(1);
  cout << ans << endl;
  return 0;
}
