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 int mod1 = 1e9 + 7, mod2 = 1e9 + 9;
int pwr(int a, long long x, int mod) {
  if (x == 0) return 1;
  int sq = pwr(a, x / 2, mod);
  sq = 1ll * sq * sq % mod;
  if (x & 1) sq = 1ll * sq * a % mod;
  return sq;
}
struct mhash {
  long long l, r;
  long long norm(long long& x, const int mod) {
    x %= mod;
    if (x < 0) {
      x += mod;
    }
    return x;
  }
  mhash(long long _l, long long _r) : l(norm(_l, mod1)), r(norm(_r, mod2)) {}
  mhash(long long x = 0) : l(norm(x, mod1)), r(norm(x, mod2)) {}
  mhash operator=(const mhash& a) {
    l = a.l, r = a.r;
    return (*this);
  }
  mhash operator+(const mhash& a) const { return mhash(l + a.l, r + a.r); }
  mhash operator+=(const mhash& a) {
    l += a.l;
    r += a.r;
    norm(l, mod1), norm(r, mod2);
    return (*this);
  }
  mhash operator*(const mhash& a) const { return mhash(l * a.l, r * a.r); }
  mhash operator*=(const mhash& a) {
    l *= a.l;
    r *= a.r;
    norm(l, mod1), norm(r, mod2);
    return (*this);
  }
  mhash operator-(const mhash& a) const { return mhash(l - a.l, r - a.r); }
  mhash operator-=(const mhash& a) {
    l -= a.l;
    r -= a.r;
    norm(l, mod1), norm(r, mod2);
    return (*this);
  }
  mhash operator/(const mhash& a) const {
    return mhash(l * pwr(a.l, mod1 - 2, mod1), r * pwr(a.r, mod2 - 2, mod2));
  }
  mhash operator/=(const mhash& a) {
    l *= pwr(a.l, mod1 - 2, mod1);
    r *= pwr(a.r, mod2 - 2, mod2);
    norm(l, mod1), norm(r, mod2);
    return (*this);
  }
  const bool operator<(const mhash& a) const {
    return l < a.l or (l == a.l and r < a.r);
  }
  const bool operator==(const mhash& a) const { return l == a.l and r == a.r; }
};
const int N = 2e5 + 10, K = 9;
vector<pair<int, int>> g[N];
mhash p[N], tot = 1, val[K][K];
int k, ans;
void calc(int i = 0, mhash h = 0) {
  if (i == k) {
    ans += tot == h;
    return;
  }
  for (int j = 0; j <= i; ++j) {
    calc(i + 1, h + val[i][j]);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m >> k;
  while (m--) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    g[u].emplace_back(w, v);
  }
  p[0] = 1;
  for (int i = 1; i < n; ++i) {
    p[i] = p[i - 1] * 2;
    tot += p[i];
  }
  for (int i = 0; i < n; ++i) {
    sort(g[i].begin(), g[i].end());
    int sz = g[i].size();
    for (int j = 0; j < sz; ++j) {
      val[sz - 1][j] += p[g[i][j].second];
    }
  }
  calc();
  cout << ans;
  return 0;
}
