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;
namespace RenaMoe {
template <typename TT>
inline void read(TT &x) {
  x = 0;
  bool f = false;
  char ch = getchar();
  while (!isdigit(ch)) f |= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x = f ? -x : x;
}
const int N = 5e5 + 9;
const int P = 998244353;
int n, k, m, ans;
int a[N], l[N], r[N], one[N], ln[N], f[N];
inline void clear() {
  for (int i = 0; i <= n + 1; ++i) ln[i] = one[i] = f[i] = 0;
}
inline int dp(int x) {
  clear();
  for (int i = 1; i <= m; ++i) {
    if ((a[i] >> x) & 1)
      one[l[i]]++, one[r[i] + 1]--;
    else
      ln[r[i] + 1] = max(ln[r[i] + 1], l[i]);
  }
  for (int i = 1; i <= n + 1; ++i) {
    one[i] += one[i - 1];
    ln[i] = max(ln[i], ln[i - 1]);
  }
  int sum = 1, p = 0;
  f[0] = 1;
  for (int i = 1; i <= n + 1; ++i) {
    while (p < ln[i]) {
      sum = (sum - f[p] + P) % P;
      f[p] = 0;
      p++;
    }
    f[i] = one[i] ? 0 : sum;
    sum = (sum + f[i]) % P;
  }
  return f[n + 1];
}
inline void main() {
  read(n), read(k), read(m);
  for (int i = 1; i <= m; ++i) read(l[i]), read(r[i]), read(a[i]);
  ans = 1;
  for (int i = 0; i < k; ++i) ans = 1ll * ans * dp(i) % P;
  printf("%d\n", ans);
}
}  // namespace RenaMoe
int main() {
  RenaMoe::main();
  return 0;
}
