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;
using ll = long long;
const int N = 2e5 + 5;
const ll base = 998244353, mod = 1e9 + 7;
struct node {
  int v, w;
} e[N][10];
int n, m, k, x, y, z, ans, a[10][N], num[N];
ll res = 1, p[N], ha[10][10];
void Dfs(int x, ll mul = 1) {
  if (x == k + 1) {
    if (mul == res) ++ans;
    return;
  }
  for (int i = 1; i <= x; ++i) Dfs(x + 1, mul * ha[x][i] % mod);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &x, &y, &z);
    e[x][++num[x]].v = y, e[x][num[x]].w = z;
  }
  p[0] = res = 1;
  for (int i = 1; i <= n; ++i) {
    sort(e[i] + 1, e[i] + 1 + num[i], [](node x, node y) { return x.w < y.w; });
    a[num[i]][++a[num[i]][0]] = i;
    p[i] = p[i - 1] * base % mod;
    (res *= i * p[i] % mod) %= mod;
  }
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j) {
      ha[i][j] = 1;
      for (int k = 1; k <= a[i][0]; ++k)
        (ha[i][j] *= e[a[i][k]][j].v * p[e[a[i][k]][j].v] % mod) %= mod;
    }
  Dfs(1);
  printf("%d\n", ans);
  return 0;
}
