You are given three integers n, k, m and m conditions (l_1, r_1, x_1), (l_2, r_2, x_2), ..., (l_m, r_m, x_m).

Calculate the number of distinct arrays a, consisting of n integers such that: 

  * 0 ≤ a_i < 2^k for each 1 ≤ i ≤ n; 
  * bitwise AND of numbers a[l_i] \& a[l_i + 1] \& ... \& a[r_i] = x_i for each 1 ≤ i ≤ m. 



Two arrays a and b are considered different if there exists such a position i that a_i ≠ b_i. 

The number can be pretty large so print it modulo 998244353.

Input

The first line contains three integers n, k and m (1 ≤ n ≤ 5 ⋅ 10^5, 1 ≤ k ≤ 30, 0 ≤ m ≤ 5 ⋅ 10^5) — the length of the array a, the value such that all numbers in a should be smaller than 2^k and the number of conditions, respectively.

Each of the next m lines contains the description of a condition l_i, r_i and x_i (1 ≤ l_i ≤ r_i ≤ n, 0 ≤ x_i < 2^k) — the borders of the condition segment and the required bitwise AND value on it.

Output

Print a single integer — the number of distinct arrays a that satisfy all the above conditions modulo 998244353.

Examples

Input


4 3 2
1 3 3
3 4 6


Output


3


Input


5 2 3
1 3 2
2 5 0
3 3 3


Output


33

Note

You can recall what is a bitwise AND operation [here](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

In the first example, the answer is the following arrays: [3, 3, 7, 6], [3, 7, 7, 6] and [7, 3, 7, 6].
#include <bits/stdc++.h>
using namespace std;
bool __hack = std::ios::sync_with_stdio(false);
auto __hack1 = cin.tie(nullptr);
template <class T>
inline bool get_bit(T mask, int bit) {
  assert(0 <= bit && bit < numeric_limits<T>::digits);
  return mask & ((T)(1) << bit);
}
namespace template_util {
constexpr int bytecount(uint64_t x) { return x ? 1 + bytecount(x >> 8) : 0; }
template <int N>
struct bytetype {};
template <uint64_t N>
struct minimal_uint : bytetype<bytecount(N)> {};
}  // namespace template_util
template <class T>
T next(istream& in) {
  T ret;
  in >> ret;
  return ret;
}
template <typename T, typename C, T Modulus>
class MR {
 private:
  struct tag_plus {};
  struct tag_minus {};
  struct tag_good {};
  T value;
  static_assert(std::numeric_limits<C>::max() / Modulus / Modulus > 0,
                "compute type is too small");
  static_assert(Modulus < std::numeric_limits<T>::max() / 2,
                "storage type is too small");
  static constexpr T reduce(T value, tag_plus) {
    return value >= Modulus ? value - Modulus : value;
  }
  static constexpr T reduce(T value, tag_minus) {
    return value < 0 ? value + Modulus : value;
  }
  static constexpr T reduce(T value, tag_good) { return value; }

 public:
  constexpr MR(C value) : value(reduce(value % Modulus, tag_minus())) {}
  template <typename tag_t>
  constexpr MR(T value, tag_t tag) : value(reduce(value, tag)) {}
  MR& operator=(C value) {
    this->value = reduce(value % Modulus, tag_minus());
    return *this;
  }
  constexpr MR operator*(MR b) const {
    return MR(C(value) * C(b.value) % Modulus, tag_good());
  }
  MR& operator+=(MR b) {
    value = reduce(value + b.value, tag_plus());
    return *this;
  }
  MR& operator-=(MR b) {
    value = reduce(value - b.value, tag_minus());
    return *this;
  }
  constexpr T get() const { return value; }
};
template <typename T, typename C, T Modulus>
static inline std::ostream& operator<<(std::ostream& o, MR<T, C, Modulus> mr) {
  return o << mr.get();
}
MR<int, int64_t, 998244353> solveOneBit(
    int n, const vector<pair<pair<int, int>, int>>& queries, int bit) {
  vector<int> sum(n + 1, 0);
  vector<int> ri(n + 1, -1);
  for (auto& p : queries) {
    if (get_bit(p.second, bit)) {
      ++sum[p.first.first];
      --sum[p.first.second];
    } else {
      ri[p.first.second] = max(ri[p.first.second], p.first.first);
    }
  }
  for (int i = 0; i < (n); ++i) {
    sum[i + 1] += sum[i];
  }
  vector<MR<int, int64_t, 998244353>> dp0(n + 1, 0);
  dp0[0] = 1;
  MR<int, int64_t, 998244353> s = 1;
  int ptr = 0;
  for (int i = 1; i <= n; ++i) {
    if (sum[i - 1] == 0) {
      dp0[i] = s;
    }
    if (ri[i] != -1) {
      while (ri[i] >= ptr) {
        s -= dp0[ptr];
        dp0[ptr] = 0;
        ++ptr;
      }
    }
    s += dp0[i];
  }
  return s;
}
void solve(istream& in, ostream& out) {
  auto n = next<int>(in);
  auto countBits = next<int>(in);
  auto m = next<int>(in);
  vector<pair<pair<int, int>, int>> queries(m);
  for (int i = 0; i < (m); ++i) {
    auto le = next<int>(in) - 1;
    auto ri = next<int>(in);
    auto x = next<int>(in);
    queries[i] = {{le, ri}, x};
  }
  MR<int, int64_t, 998244353> res = 1;
  for (int bit = 0; bit < (countBits); ++bit) {
    res = res * solveOneBit(n, queries, bit);
  }
  out << res << "\n";
}
int main() {
  solve(cin, cout);
  return 0;
}
