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;
vector<int> u[500008], v[500008];
int n, k, m, l, r, x, z = 998244353, a[500008], d[500008];
int main() {
  for (scanf("%d%d%d", &n, &k, &m); m--;
       scanf("%d%d%d", &l, &r, &x), u[l].push_back(r), v[l].push_back(x))
    ;
  for (d[0] = m = 1; k-- && m; m = m * (0LL | d[n + 1] - d[n]) % z) {
    for (l = n + 2; l--; a[l] = -1)
      ;
    for (l = r = 0; l++ <= n;
         d[l] =
             (d[l - 1] + (r < l ? d[l - 1] - (~a[l] ? d[a[l]] : 0) : 0) % z) %
             z)
      for (a[l] < a[l - 1] ? a[l] = a[l - 1] : 0, x = v[l].size(); x--;
           1 << k & v[l][x] ? r = max(r, u[l][x])
                            : a[u[l][x] + 1] = max(a[u[l][x] + 1], l - 1))
        ;
  }
  printf("%d\n", m < 0 ? m + z : m), exit(0);
}
