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;
struct query {
  int l, r, x;
} q[505050];
const long long mod = 998244353;
long long reserve, pre[505050];
int a[505050], lst[505050];
long long solve(int p) {
  memset(pre, 0, sizeof pre);
  memset(a, 0, sizeof a);
  memset(lst, 0, sizeof lst);
  for (int i = 1; i <= m; ++i) {
    if ((q[i].x >> p) & 1) {
      a[q[i].l]++;
      a[q[i].r + 1]--;
    } else {
      lst[q[i].r] = max(lst[q[i].r], q[i].l);
    }
  }
  for (int i = 1; i <= n + 1; ++i) a[i] += a[i - 1];
  pre[0] = 0;
  int beg = 0;
  for (int i = 1; i <= n + 1; ++i) {
    if (a[i]) {
      pre[i] = pre[i - 1];
    } else {
      if (!beg)
        pre[i] = pre[i - 1] + pre[i - 1] + 1;
      else
        pre[i] = pre[i - 1] + pre[i - 1] - pre[beg - 1] + mod;
    }
    pre[i] %= mod;
    beg = max(beg, lst[i]);
  }
  return (pre[n + 1] - pre[n] + mod) % mod;
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; ++i) scanf("%d%d%d", &q[i].l, &q[i].r, &q[i].x);
  long long ans = 1;
  for (int i = 0; i < k; ++i) {
    (ans *= solve(i)) %= mod;
  }
  cout << ans << endl;
}
