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;
using ll = long long;
using ii = pair<int, int>;
const int N = 500005;
const int mod = 998244353;
int psum[N], t1[N];
ll ssum[N];
ll dp[N];
int main() {
  int p2[N];
  p2[0] = 1;
  for (int i = 1; i < N; i++) {
    p2[i] = (p2[i - 1] * 2) % mod;
  }
  int n, k, m;
  scanf("%d %d %d", &n, &k, &m);
  vector<pair<ii, int>> v;
  for (int i = 0; i < m; i++) {
    int l, r, x;
    scanf("%d %d %d", &l, &r, &x);
    v.push_back({{l, r}, x});
  }
  sort(v.begin(), v.end());
  ll ans = 1;
  for (int b = 0; b < k; b++) {
    int id = 0;
    for (int i = 1; i <= n; i++) {
      while (id < m && (!(v[id].second & (1 << b)) || v[id].first.second < i)) {
        id++;
      }
      if (id < m && v[id].first.first <= i) {
        t1[i] = 1;
      } else {
        t1[i] = 0;
      }
      psum[i] = psum[i - 1] + t1[i];
    }
    id = m - 1;
    dp[n + 1] = ssum[n + 1] = 1;
    int nl = n + 1, nr = n + 1;
    for (int i = n; i >= 0; i--) {
      while (id >= 0 && v[id].first.first > i) {
        while (id >= 0 && (v[id].second & (1 << b))) {
          id--;
        }
        while (id >= 0 && !(v[id].second & (1 << b)) && v[id].first.first > i) {
          nl = min(nl, v[id].first.first);
          nr = min(nr, v[id].first.second);
          id--;
        }
      }
      if (t1[i]) {
        dp[i] = 0;
      } else {
        int q2 = (nl - i - 1) - (psum[nl - 1] - psum[i]);
        dp[i] = (ssum[nl] - ssum[nr + 1] + mod) % mod;
        dp[i] *= p2[q2];
        dp[i] %= mod;
      }
      ssum[i] = ssum[i + 1] + dp[i];
      ssum[i] %= mod;
    }
    ans *= dp[0];
    ans %= mod;
  }
  printf("%lld\n", ans);
  return 0;
}
