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;
template <typename T>
using Vec = std::vector<T>;
const int MOD = 998244353;
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  int n, k, m;
  cin >> n >> k >> m;
  Vec<int> l(m), r(m), x(m);
  for (int i = 0; i < (m); ++i) {
    cin >> l[i] >> r[i] >> x[i];
  }
  int ans = 1;
  for (int i = 0; i < (k); ++i) {
    Vec<int> prefix_sum(n + 2, 0);
    Vec<int> max_l(n + 2, 0);
    Vec<int> dp(n + 2, 0);
    for (int j = 0; j < (m); ++j) {
      if (x[j] >> i & 1) {
        prefix_sum[l[j]] += 1;
        prefix_sum[r[j] + 1] -= 1;
      } else {
        max_l[r[j]] = std::max(max_l[r[j]], l[j]);
      }
    }
    for (int j = (1); j < (n + 2); ++j) {
      prefix_sum[j] += prefix_sum[j - 1];
    }
    for (int j = (1); j < (n + 1); ++j) {
      max_l[j] = std::max(max_l[j], max_l[j - 1]);
    }
    dp[0] = 1;
    for (int j = (1); j < (n + 2); ++j) {
      if (prefix_sum[j] == 0) {
        if (max_l[j - 1] == 0) {
          dp[j] = (2LL * dp[j - 1]) % MOD;
        } else {
          dp[j] = (2LL * dp[j - 1] - dp[max_l[j - 1] - 1] + MOD) % MOD;
        }
      } else {
        dp[j] = dp[j - 1];
      }
    }
    dp[n + 1] = (dp[n + 1] - dp[n] + MOD) % MOD;
    ans = (1LL * ans * dp[n + 1]) % MOD;
  }
  cout << ans % MOD << endl;
  return 0;
}
