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;
struct event {
  long long x;
  bool start;
  long long ot;
  long long v;
  event(long long ix, long long io, long long iv, bool ist)
      : x(ix), ot(io), v(iv), start(ist) {}
};
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, k;
  cin >> n >> k >> m;
  vector<vector<event>> evs(n + 2);
  for (long long i = 0; i < m; i++) {
    long long l, r, v;
    cin >> l >> r >> v;
    evs[l].emplace_back(l, r, v, true);
    evs[r + 1].emplace_back(r, l, v, false);
  }
  long long res = 1;
  for (long long i = 0; i < k; i++) {
    long long bit = 1ll << i;
    vector<long long> dp(n + 2);
    dp[0] = 1;
    long long one = 0;
    long long mmax = 0;
    vector<long long> prefsum(n + 2, 0);
    for (long long i = 1; i <= n + 1; i++) {
      prefsum[i] = prefsum[i - 1] + dp[i - 1];
      for (auto e : evs[i]) {
        if (e.v & bit) {
          if (e.start)
            one++;
          else
            one--;
        } else {
          if (!e.start) {
            mmax = max(mmax, e.ot);
          }
        }
      }
      if (one == 0) dp[i] += prefsum[i] - prefsum[mmax];
      dp[i] = (dp[i] + 4 * 998244353ll) % 998244353ll;
      prefsum[i] = (prefsum[i] + 4 * 998244353ll) % 998244353ll;
    }
    res = res * dp[n + 1];
    res %= 998244353ll;
  }
  cout << res << "\n";
}
