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 N = 2e5 + 10, mod = 1e9 + 7;
pair<long long, long long> operator+(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  return make_pair((a.first + b.first) % mod, (a.second + b.second) % mod);
}
pair<long long, long long> operator+(pair<long long, long long> a,
                                     long long b) {
  return make_pair((a.first * b) % mod, (a.second * b) % mod);
}
pair<long long, long long> operator-(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  return make_pair((a.first - b.first + mod) % mod,
                   (a.second - b.second + mod) % mod);
}
pair<long long, long long> operator*(pair<long long, long long> a,
                                     pair<long long, long long> b) {
  return make_pair((a.first * b.first) % mod, (a.second * b.second) % mod);
}
struct edge {
  int y, z;
  bool operator<(const edge &a) const { return z < a.z; }
} e[N];
int n, m, K, ans;
pair<long long, long long> bas = make_pair(131, 13331), h[N], tot, pre[15][15];
vector<edge> g[N];
void dfs(int x, pair<long long, long long> cur) {
  if (x == K + 1) {
    ans += (cur == tot);
    return;
  }
  for (int i = 0; i <= x - 1; i++) dfs(x + 1, cur + pre[x][i]);
}
int main() {
  cin >> n >> m >> K;
  h[0] = make_pair(1, 1);
  for (int i = 1; i <= n; i++) h[i] = h[i - 1] * bas, tot = tot + h[i];
  for (int i = 1; i <= m; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    g[x].push_back((edge){y, z});
  }
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end());
    int sz = g[i].size();
    for (int j = 0; j <= sz - 1; j++) pre[sz][j] = pre[sz][j] + h[g[i][j].y];
  }
  dfs(1, make_pair(0, 0));
  printf("%d\n", ans);
  return 0;
}
