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 Info {
  int v, w;
} Edge[200007];
long long const mod[3] = {1000000007, 1226959, 998244353};
int n, m, k, c[200007], e[200007][10], randnum[200007][3], hashans[3];
int f[10][10][3];
bool cmp(int a, int b) { return Edge[a].w < Edge[b].w; }
int ans;
void dfs(int ste, int s0, int s1, int s2) {
  if (ste > k) {
    if (s0 == hashans[0] && s1 == hashans[1] && s2 == hashans[2]) ++ans;
    return;
  }
  for (int i = 1; i <= ste; ++i)
    dfs(ste + 1, 1ll * s0 * f[ste][i][0] % mod[0],
        1ll * s1 * f[ste][i][1] % mod[1], (s2 + f[ste][i][2]) % mod[2]);
}
int main() {
  srand((unsigned)time(NULL));
  scanf("%d%d%d", &n, &m, &k);
  hashans[0] = hashans[1] = 1;
  for (int i = 1; i <= n; ++i) {
    randnum[i][0] = abs(rand() % mod[0]), randnum[i][1] = abs(rand() % mod[1]);
    randnum[i][2] = abs(rand() % mod[2]);
    hashans[0] = 1ll * hashans[0] * randnum[i][0] % mod[0];
    hashans[1] = 1ll * hashans[1] * randnum[i][1] % mod[1];
    hashans[2] = (hashans[2] + randnum[i][2]) % mod[2];
  }
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    Edge[i] = (Info){v, w};
    e[u][++c[u]] = i;
  }
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j) f[i][j][0] = f[i][j][1] = 1;
  for (int i = 1; i <= n; ++i) {
    sort(e[i] + 1, e[i] + c[i] + 1, cmp);
    for (int j = 1; j <= c[i]; ++j) {
      f[c[i]][j][0] =
          1ll * f[c[i]][j][0] * randnum[Edge[e[i][j]].v][0] % mod[0];
      f[c[i]][j][1] =
          1ll * f[c[i]][j][1] * randnum[Edge[e[i][j]].v][1] % mod[1];
      f[c[i]][j][2] = (f[c[i]][j][2] + randnum[Edge[e[i][j]].v][2]) % mod[2];
    }
  }
  dfs(1, 1, 1, 0);
  printf("%d\n", ans);
  return 0;
}
