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 mo = 998244353;
int n, k, m, ans = 1;
int f[500005], sum[500005];
int l[500005], r[500005], x[500005];
int tend[500005], cf[500005];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &l[i], &r[i], &x[i]);
  }
  for (int i = 0; i < k; i++) {
    memset(tend, 0, sizeof(tend));
    memset(cf, 0, sizeof(cf));
    for (int j = 1; j <= m; j++) {
      if ((x[j] >> i) & 1) {
        cf[l[j]]++;
        cf[r[j] + 1]--;
      } else
        tend[r[j]] = max(tend[r[j]], l[j]);
    }
    int tmp = 0, last = 0;
    f[0] = sum[0] = 1;
    for (int j = 1; j <= n + 1; j++) {
      tmp += cf[j];
      if (tmp > 0)
        f[j] = 0;
      else
        f[j] = last == 0 ? sum[j - 1] : (sum[j - 1] - sum[last - 1] + mo) % mo;
      sum[j] = (sum[j - 1] + f[j]) % mo;
      last = max(last, tend[j]);
    }
    ans = (long long)ans * f[n + 1] % mo;
  }
  printf("%d", ans);
  return 0;
}
