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 T>
inline void read(T &x) {
  int f = 0, c = getchar();
  x = 0;
  while (!isdigit(c)) f |= c == 45, c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  if (f) x = -x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &...args) {
  read(x);
  read(args...);
}
const int maxn = 5e5 + 50, mod = 998244353;
int n, K, m, a[maxn];
long long f[maxn], ans = 1;
int pos[maxn], l[maxn], r[maxn], x[maxn];
void init(int bi) {
  for (int i = 1; i <= n + 1; i++) pos[i] = a[i] = 0;
  for (int i = 1; i <= m; i++)
    if (x[i] & (1 << bi))
      a[l[i]]++, a[r[i] + 1]--;
    else
      pos[r[i] + 1] = max(pos[r[i] + 1], l[i]);
  for (int i = 2; i <= n + 1; i++)
    a[i] = a[i - 1] + a[i], pos[i] = max(pos[i], pos[i - 1]);
}
void dp() {
  for (int i = 1; i <= n + 1; i++) f[i] = 0;
  f[0] = 1;
  long long sum = 1, pt = 0;
  for (int i = 1; i <= n + 1; i++) {
    while (pt < pos[i]) sum = (sum - f[pt] + mod) % mod, pt++;
    if (a[i] > 0)
      f[i] = 0;
    else
      f[i] = sum, sum += f[i], sum %= mod;
  }
}
signed main() {
  read(n, K, m);
  for (int i = 1; i <= m; i++) read(l[i], r[i], x[i]);
  for (int i = 0; i < K; i++) {
    init(i);
    dp();
    ans *= f[n + 1];
    ans %= mod;
  }
  cout << ans;
  return 0;
}
