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;
vector<pair<int, int> > v[200005];
vector<pair<int, int> > in[200005];
long long go[10][10][10][10];
int pre[200005][10][10];
int suf[200005][10][10];
long long ans = 0;
long long ok[10][10];
vector<int> res;
void dfs(int first) {
  if (first == 0) {
    ans++;
    return;
  }
  for (int i = 0; i < first; i++) {
    for (int a = 0; a < 10; a++) {
      for (int b = 0; b < 10; b++) {
        ok[a][b] -= go[first][i][a][b];
      }
    }
    if (ok[first][i] == 1) {
      dfs(first - 1);
    }
    for (int a = 0; a < 10; a++) {
      for (int b = 0; b < 10; b++) {
        ok[a][b] += go[first][i][a][b];
      }
    }
  }
}
void solve() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) ok[i][j] = 1;
  }
  for (int i = 0; i < m; i++) {
    int first, second, w;
    scanf("%d %d %d", &first, &second, &w);
    v[first].push_back(make_pair(w, second));
  }
  for (int i = 1; i <= n; i++) {
    sort(v[i].begin(), v[i].end());
    for (int j = 0; j < v[i].size(); j++) {
      in[v[i][j].second].push_back(make_pair(v[i].size(), j));
    }
  }
  for (int i = 1; i <= n; i++) {
    memset(pre[0], 0, sizeof(pre[0]));
    for (int j = 0; j < in[i].size(); j++) {
      for (int a = 0; a < 10; a++) {
        for (int b = 0; b < 10; b++) {
          pre[j + 1][a][b] = pre[j][a][b];
        }
      }
      pre[j + 1][in[i][j].first][in[i][j].second] = 1;
    }
    memset(suf[in[i].size()], 0, sizeof(suf[in[i].size()]));
    for (int j = in[i].size() - 1; j >= 0; j--) {
      for (int a = 0; a < 10; a++) {
        for (int b = 0; b < 10; b++) {
          suf[j][a][b] = suf[j + 1][a][b];
        }
      }
      suf[j][in[i][j].first][in[i][j].second] = 1;
    }
    for (int j = 0; j < in[i].size(); j++) {
      for (int a = 0; a < 10; a++) {
        for (int b = 0; b < 10; b++) {
          go[in[i][j].first][in[i][j].second][a][b] |=
              pre[j][a][b] | suf[j + 1][a][b];
        }
      }
    }
  }
  dfs(k);
  printf("%lld\n", ans);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
}
