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 pii = pair<int, int>;
const int maxn = 5e5 + 10, mod = 998244353;
int l[maxn], r[maxn], v[maxn], n, k, m;
int solve(int x) {
  vector<int> dp(n + 5), vec(n + 5), ps(n + 5), event(n + 5), mark(m + 5);
  for (int i = 0; i < m; i++) {
    if (v[i] >> x & 1) {
      ++vec[l[i]];
      --vec[r[i] + 1];
    } else {
      event[r[i] + 1] = max(event[r[i] + 1], l[i]);
    }
  }
  for (int i = 1; i <= n; i++) vec[i] += vec[i - 1];
  dp[0] = 1, ps[0] = 1;
  int last = 0;
  for (int i = 1; i <= n + 1; i++) {
    last = max(last, event[i]);
    ps[i] = ps[i - 1];
    if (vec[i]) continue;
    (dp[i] = ps[i - 1] + (last ? mod - ps[last - 1] : 0)) %= mod;
    (ps[i] += dp[i]) %= mod;
  }
  return (ps[n] + (last ? mod - ps[last - 1] : 0)) % mod;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> m;
  for (int i = 0; i < m; i++) cin >> l[i] >> r[i] >> v[i];
  int ans = 1;
  for (int i = 0; i < k; i++) (ans = 1ll * ans * solve(i) % mod) %= mod;
  cout << ans << '\n';
  return 0;
}
