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;
int n, k, m;
pair<pair<int, int>, int> nr[500005];
int s[500005], maxi[500005], dp[500005];
int sum(int x, int y) {
  x += y;
  if (x < 0) x += 998244353;
  if (x >= 998244353) x -= 998244353;
  return x;
}
int main() {
  int i;
  scanf("%d%d%d", &n, &k, &m);
  for (i = 1; i <= m; i++)
    scanf("%d%d%d", &nr[i].first.first, &nr[i].first.second, &nr[i].second);
  int rasp = 1, st, dr, par, i1;
  for (i1 = 0; i1 < k; i1++) {
    memset(s, 0, sizeof(s));
    memset(maxi, -1, sizeof(maxi));
    for (i = 1; i <= m; i++) {
      st = nr[i].first.first;
      dr = nr[i].first.second;
      if ((nr[i].second & (1 << i1)) == 0) maxi[dr] = max(maxi[dr], st);
      if (nr[i].second & (1 << i1)) {
        s[st]++;
        s[dr + 1]--;
      }
    }
    int poz = -1;
    for (i = 2; i <= n; i++) s[i] = s[i] + s[i - 1];
    for (i = 1; i <= n; i++) {
      par = 0;
      if (s[i] == 0) {
        par = dp[i];
        if (poz == -1)
          par = sum(par, 1);
        else
          par = sum(par, -dp[poz]);
      }
      dp[i + 1] = sum(dp[i], par);
      poz = max(poz, maxi[i]);
    }
    if (poz == -1)
      rasp = 1LL * rasp * sum(dp[n + 1], 1) % 998244353;
    else
      rasp = 1LL * rasp * sum(dp[n + 1], -dp[poz]) % 998244353;
  }
  printf("%d\n", rasp);
  return 0;
}
