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;
inline int read() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + (ch ^ '0');
    ch = getchar();
  }
  return f ? x : -x;
}
const int N = 5e5 + 5, mod = 998244353;
int n, k, m, l[N], r[N], x[N], c[N], lim[N], f[N];
inline void chkmax(int& x, int y) { x = x > y ? x : y; }
inline void inc(int& x, int y) { x = x + y < mod ? x + y : x + y - mod; }
inline void dec(int& x, int y) { x = x - y >= 0 ? x - y : x - y + mod; }
int work(int k) {
  memset(c, 0, sizeof c), memset(lim, 0, sizeof lim);
  for (int i = (1), _ed = (m); i <= _ed; ++i)
    if (x[i] >> k & 1)
      ++c[l[i]], --c[r[i] + 1];
    else
      chkmax(lim[r[i]], l[i]);
  for (int i = (1), _ed = (n + 1); i <= _ed; ++i)
    c[i] += c[i - 1], chkmax(lim[i], lim[i - 1]);
  f[0] = 1;
  for (int i = (1), _ed = (n + 1); i <= _ed; ++i) {
    f[i] = f[i - 1];
    if (c[i]) continue;
    if (lim[i - 1]) dec(f[i], f[lim[i - 1] - 1]);
    if (i <= n) inc(f[i], f[i - 1]);
  }
  return f[n + 1];
}
int main() {
  n = read(), k = read(), m = read();
  for (int i = (1), _ed = (m); i <= _ed; ++i)
    l[i] = read(), r[i] = read(), x[i] = read();
  int ans = 1;
  for (int i = (0), _ed = (k - 1); i <= _ed; ++i)
    ans = 1ll * ans * work(i) % mod;
  printf("%d\n", ans);
  return 0;
}
