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;
const long long N = 2e5 + 7, K = 10;
long long n, m, k;
vector<pair<long long, long long> > g[N];
bitset<N> who[K][K];
long long cnt[N];
bool bad[K][K];
long long ban[K][K], num[K][K];
long long ans = 0;
void solve(long long i, long long b) {
  if (i == k + 1) {
    ans++;
    return;
  }
  for (long long j = 1; j <= i; ++j) {
    if (!bad[i][j] && !((b >> num[i][j]) & 1)) {
      solve(i + 1, b | ban[i][j]);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (long long i = 0; i < m; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    g[u].push_back(make_pair(w, v));
  }
  for (long long u = 1; u <= n; ++u) sort(g[u].begin(), g[u].end());
  for (long long x = 1; x <= k; ++x) {
    for (long long y = 1; y <= x; ++y) {
      memset(cnt, 0, sizeof cnt);
      for (long long u = 1; u <= n; ++u) {
        if (g[u].size() == x) {
          long long v = g[u][y - 1].second;
          cnt[v]++;
        }
      }
      for (long long u = 1; u <= n; ++u) {
        if (cnt[u] > 1) {
          bad[x][y] = 1;
          break;
        }
      }
      for (long long u = 1; u <= n; ++u) {
        if (cnt[u]) {
          who[x][y][u] = 1;
        }
      }
    }
  }
  long long ptr = 0;
  for (long long i = 1; i <= k; ++i)
    for (long long j = 1; j <= i; ++j) num[i][j] = ptr++;
  for (long long x1 = 1; x1 <= k; ++x1) {
    for (long long y1 = 1; y1 <= x1; ++y1) {
      for (long long x2 = x1 + 1; x2 <= k; ++x2) {
        for (long long y2 = 1; y2 <= x2; ++y2) {
          if (!bad[x1][y1] && !bad[x2][y2]) {
            if ((who[x1][y1] & who[x2][y2]).count()) {
              ban[x1][y1] += 1ll << num[x2][y2];
            }
          }
        }
      }
    }
  }
  solve(1, 0);
  cout << ans << '\n';
}
