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 mod = 998244353;
const int maxn = 5e5 + 10;
int n, m, k;
int l[maxn], r[maxn], a[maxn], b[maxn];
int cl[maxn], dif[maxn];
long long f[maxn], psum[maxn];
long long solve() {
  for (int i = 1; i <= n; i++) dif[i] = 0, cl[i] = -1;
  for (int i = 1; i <= m; i++) {
    if (b[i])
      ++dif[l[i]], --dif[r[i] + 1];
    else
      cl[r[i]] = max(cl[r[i]], l[i]);
  }
  int leftmost = 0;
  f[0] = psum[0] = 1;
  for (int i = 1; i <= n; i++) {
    dif[i] += dif[i - 1];
    if (dif[i])
      f[i] = 0;
    else {
      f[i] = psum[i - 1];
      if (leftmost) f[i] = (f[i] - psum[leftmost - 1] + mod) % mod;
    }
    psum[i] = (f[i] + psum[i - 1]) % mod;
    leftmost = max(leftmost, cl[i]);
  }
  long long res = 0;
  for (int i = leftmost; i <= n; i++) res = (res + f[i]) % mod;
  return res;
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &l[i], &r[i], &a[i]);
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 1; j <= m; j++) b[j] = a[j] >> i & 1;
    ans = ans * solve() % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
