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 MAXN = 500005;
const int MOD = 998244353;
int n, k, m;
inline int modmul(int a, int b) { return (a * 1ll * b) % MOD; }
inline int modadd(int a, int b) { return (a + b) % MOD; }
pair<pair<int, int>, int> ranges[MAXN];
int solve_for_bit(int bit_id) {
  vector<int> cnt_set_bits_on_range(n + 2, 0);
  vector<int> max_l_ending_at(n + 2, 0);
  vector<int> g(n + 2, 0);
  for (int i = 1; i <= m; ++i) {
    int l = ranges[i].first.first;
    int r = ranges[i].first.second;
    bool is_set = ((ranges[i].second & (1 << bit_id)) != 0);
    if (is_set) {
      cnt_set_bits_on_range[l]++;
      cnt_set_bits_on_range[r + 1]--;
    } else {
      max_l_ending_at[r] = max(max_l_ending_at[r], l);
    }
  }
  int cur_set_bits = 0;
  int cur_g = 0;
  for (int i = 1; i <= n + 1; ++i) {
    cur_set_bits += cnt_set_bits_on_range[i];
    cnt_set_bits_on_range[i] = (cur_set_bits > 0);
    g[i] = cur_g;
    cur_g = max(cur_g, max_l_ending_at[i]);
  }
  vector<int> f(n + 2, 1);
  vector<int> pref_f(n + 2, 1);
  for (int i = 1; i <= n + 1; ++i) {
    if (cnt_set_bits_on_range[i]) {
      f[i] = 0;
    } else {
      f[i] = (pref_f[i - 1] - (g[i] ? pref_f[g[i] - 1] : 0) + MOD) % MOD;
    }
    pref_f[i] = modadd(pref_f[i - 1], f[i]);
  }
  return f[n + 1];
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &ranges[i].first.first, &ranges[i].first.second,
          &ranges[i].second);
  }
  ranges[0] = {{0, 0}, 0};
  int ans = 1;
  for (int i = 0; i < k; ++i) {
    ans = modmul(ans, solve_for_bit(i));
  }
  printf("%d\n", ans);
  return 0;
}
