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;
template <typename Arg1>
void prn(Arg1&& arg1) {
  cout << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void prn(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prn(args...);
}
template <typename Arg1>
void prs(Arg1&& arg1) {
  cout << arg1 << " ";
}
template <typename Arg1, typename... Args>
void prs(Arg1&& arg1, Args&&... args) {
  cout << arg1 << " ";
  prs(args...);
}
template <typename Arg1>
void read(Arg1&& arg1) {
  cin >> arg1;
}
template <typename Arg1, typename... Args>
void read(Arg1&& arg1, Args&&... args) {
  cin >> arg1;
  read(args...);
}
const int N = 5e5 + 5, K = 30, mod = 998244353;
int n, k, m;
vector<vector<int> > qry;
inline void add(int& a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
inline void sub(int& a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
int one[N], cdp[N], mx[N];
inline int cal(int bit) {
  for (int i = 1; i <= n; ++i) cdp[i] = one[i] = mx[i] = 0;
  for (auto& i : qry) {
    if ((i[2] >> bit) & 1) {
      ++one[i[0]];
      --one[i[1] + 1];
    } else
      mx[i[1]] = max(mx[i[1]], i[0]);
  }
  for (int i = 1; i <= n; ++i) one[i] += one[i - 1];
  int dp, mxl = 0;
  cdp[0] = 1;
  for (int i = 1; i <= n; ++i) {
    cdp[i] = cdp[i - 1];
    if (!one[i]) {
      dp = cdp[i];
      if (mxl > 0) sub(dp, cdp[mxl - 1]);
      add(cdp[i], dp);
    }
    mxl = max(mxl, mx[i]);
  }
  int ans = cdp[n];
  if (mxl > 0) sub(ans, cdp[mxl - 1]);
  return ans;
}
void solve() {
  read(n, k, m);
  for (int i = 0; i < m; ++i) {
    int l, r, x;
    read(l, r, x);
    qry.push_back({l, r, x});
  }
  long long ans = 1;
  for (int bit = 0; bit < k; ++bit) {
    ans = ans * cal(bit) % mod;
  }
  prn(ans);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (int tc = 1; tc <= t; ++tc) {
    solve();
  }
  return 0;
}
