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;
struct frez {
  int u, v, w, r, in;
} a[200050];
bool cmp1(frez a, frez b) { return a.u == b.u ? a.w < b.w : a.u < b.u; }
bool cmp2(frez a, frez b) { return a.v < b.v; }
int n, m, K, ans, cnt[15][15], c[20], deg[200050];
bool vis[15][15][15][15];
void dfs(int x) {
  if (x == K + 1) {
    ans++;
    return;
  }
  for (int i = 1; i <= x; i++) {
    bool pd = 1;
    c[x] = i;
    for (int j = 1; j <= x; j++)
      if (!vis[j][c[j]][x][i]) pd = 0;
    if (!pd) continue;
    dfs(x + 1);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &K);
  for (int i = 1; i <= m; i++)
    scanf("%d%d%d", &a[i].u, &a[i].v, &a[i].w), deg[a[i].v]++;
  for (int i = 1; i <= n; i++)
    if (!deg[i]) {
      printf("0\n");
      return 0;
    }
  sort(a + 1, a + m + 1, cmp1);
  for (int i = 1, j = 1; i <= m; i = j) {
    for (; a[i].u == a[j].u && j <= m; j++)
      ;
    for (int k = i; k < j; k++) a[k].r = k - i + 1, a[k].in = j - i;
  }
  sort(a + 1, a + m + 1, cmp2);
  for (int aa = 1; aa <= 9; aa++)
    for (int bb = 1; bb <= 9; bb++)
      for (int cc = 1; cc <= 9; cc++)
        for (int dd = 1; dd <= 9; dd++) vis[aa][bb][cc][dd] = 1;
  for (int i = 1, j = 1; i <= m; i++) {
    memset(cnt, 0, sizeof(cnt));
    for (; a[i].v == a[j].v && j <= m; j++) cnt[a[j].in][a[j].r]++;
    for (int aa = 1; aa <= 9; aa++)
      for (int bb = 1; bb <= aa; bb++)
        if (cnt[aa][bb])
          for (int cc = 1; cc <= 9; cc++)
            for (int dd = 1; dd <= cc; dd++)
              if (cnt[cc][dd]) {
                if (aa == cc && bb == dd && cnt[aa][bb] == 1) continue;
                vis[aa][bb][cc][dd] = 0;
              }
  }
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
