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;
long long int const mod = 998244353;
long long int power(long long int x, long long int y, long long int m) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2, m) % m;
  if (y % 2 == 0)
    return ((temp) * (temp)) % m;
  else
    return (((x) % m) * ((temp * temp) % m)) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    long long int n, k, m;
    cin >> n >> k >> m;
    long long int ans = 1;
    vector<pair<pair<long long int, long long int>, long long int> > vec(m);
    for (int i = 0; i < (int)m; i++) {
      cin >> vec[i].first.first >> vec[i].first.second >> vec[i].second;
    }
    for (int i = 0; i < (int)k; i++) {
      vector<long long int> temp(n + 2, 0);
      vector<long long int> ind(n + 2, -1);
      vector<long long int> dp(n + 2, 0);
      dp[0] = 1;
      for (auto j : vec) {
        if ((j.second >> i) & 1) {
          temp[j.first.first]++;
          temp[j.first.second + 1]--;
        } else {
          ind[j.first.second] = max(ind[j.first.second], j.first.first - 1);
        }
      }
      long long int id = -1;
      long long int sum = 0;
      for (int j = 1; j <= n; j++) {
        sum += temp[j];
        dp[j] = dp[j - 1];
        if (sum == 0) {
          if (id >= 0)
            dp[j] = (dp[j] + dp[j - 1] - dp[id] + mod) % mod;
          else {
            dp[j] = (dp[j] + dp[j - 1]) % mod;
          }
        }
        id = max(ind[j], id);
      }
      long long int num = dp[n];
      if (id >= 0) num = (num - dp[id] + mod) % mod;
      ans = (ans * num) % mod;
    }
    cout << ans << "\n";
  }
}
