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;
const int N = 5e5 + 5, mod = 998244353;
int n, k, m, l[N], r[N], x[N], ans = 1, p, f[N], pos[N], a[N], sum;
int read() {
  int x = 0, f = 1;
  char s;
  while ((s = getchar()) > '9' || s < '0')
    if (s == '-') f = -1;
  while (s >= '0' && s <= '9')
    x = (x << 1) + (x << 3) + (s ^ 48), s = getchar();
  return x * f;
}
void inc(int &x, const int y) { x = (x + y >= mod ? x + y - mod : x + y); }
void dec(int &x, const int y) { x = (x < y ? x - y + mod : x - y); }
int main() {
  n = read(), k = read(), m = read();
  for (int i = 1; i <= m; ++i) l[i] = read(), r[i] = read() + 1, x[i] = read();
  for (int i = 0; i < k; ++i) {
    for (int j = 1; j <= n + 1; ++j) a[j] = pos[j] = 0;
    for (int j = 1; j <= m; ++j)
      if (x[j] >> i & 1)
        ++a[l[j]], --a[r[j]];
      else
        pos[r[j]] = max(pos[r[j]], l[j]);
    for (int j = 2; j <= n + 1; ++j)
      a[j] += a[j - 1], pos[j] = max(pos[j], pos[j - 1]), f[j] = 0;
    sum = 1;
    f[1] = 0;
    p = 0;
    f[0] = 1;
    for (int j = 1; j <= n + 1; ++j) {
      while (p < pos[j]) dec(sum, f[p]), f[p] = 0, ++p;
      f[j] = (a[j] ? 0 : sum);
      inc(sum, f[j]);
    }
    ans = 1ll * ans * f[n + 1] % mod;
  }
  printf("%d\n", ans);
  return 0;
}
