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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const long long N = 2e5 + 7;
const long long M = 1e9 + 7;
const long long MAXN = 2e18 + 7;
const long long Mod = 998244353;
long long _, i, j, k, n, m, p, q, s, T, t, l, r, o, u, v, w, x, y, z, ans, nex,
    num, st, en, sx, sy, tx, ty, len, th, ma, mi, mod, sum, res, flag, cas, ret,
    mid, cnt, now, tmp;
long long a[N], b[N], c[10][10][10][10], d[N], mp[10][10];
char ch;
vector<pair<long long, long long> > g[N], h[N], f;
string s1, s2, s3, ss;
void dfs(long long cnt) {
  if (cnt == k + 1) {
    ans++;
    return;
  }
  long long i;
  for (long long x = 0; x < cnt; x++) {
    d[cnt] = x;
    for (i = 1; i <= cnt; i++)
      if (c[i][d[i]][cnt][x]) break;
    if (i <= cnt) continue;
    dfs(cnt + 1);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  long long T = 1;
  while (T--) {
    n = read();
    m = read();
    k = read();
    for (i = 1; i <= m; i++) {
      u = read();
      v = read();
      w = read();
      g[u].push_back(pair<long long, long long>(w, v));
      d[u]++;
      b[v]++;
    }
    for (i = 1; i <= n; i++) {
      if (d[i] == 0 || b[i] == 0) break;
    }
    if (i <= n) {
      printf("0\n");
      continue;
    }
    for (i = 1; i <= n; i++) sort((g[i]).begin(), (g[i]).end());
    for (i = 1; i <= n; i++) {
      for (j = 0; j < g[i].size(); j++) {
        v = g[i][j].second;
        h[v].push_back(pair<long long, long long>(d[i], j));
      }
    }
    for (i = 1; i <= n; i++) {
      sort((h[i]).begin(), (h[i]).end());
      for (l = 1; l <= k; l++)
        for (j = 0; j < l; j++) mp[l][j] = 0;
      for (j = 0; j < h[i].size(); j++) {
        if (mp[h[i][j].first][h[i][j].second])
          c[h[i][j].first][h[i][j].second][h[i][j].first][h[i][j].second] = 1;
        mp[h[i][j].first][h[i][j].second] = 1;
      }
      f.clear();
      for (l = 1; l <= k; l++)
        for (j = 0; j < l; j++)
          if (mp[l][j]) f.push_back(pair<long long, long long>(l, j));
      for (l = 0; l < f.size(); l++)
        for (j = l + 1; j < f.size(); j++) {
          c[f[l].first][f[l].second][f[j].first][f[j].second] = 1;
        }
    }
    ans = 0;
    dfs(1);
    printf("%lld\n", ans);
  }
  return 0;
}
