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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const int MOD = 998244353;
  int n, k, m;
  while (cin >> n >> k >> m) {
    vector<vector<tuple<int, int, int>>> events(n + 2);
    for (int i = 0; i < m; ++i) {
      int l, r, x;
      cin >> l >> r >> x;
      events[l].emplace_back(0, x, i);
      events[r + 1].emplace_back(l, x, i);
    }
    ll res = 1;
    for (int bit = 1; bit < (1 << k); bit <<= 1) {
      int cnt1 = 0;
      vector<char> ones(n + 2);
      for (int i = 1; i <= n; ++i) {
        for (auto [open, x, idx] : events[i]) {
          if (x & bit) {
            cnt1 += (open ? -1 : +1);
          }
        }
        if (cnt1) {
          ones[i] = true;
        }
      }
      int right_start = 0;
      vector<int> dp(n + 2);
      vector<int> pref(n + 3);
      dp[0] = 1;
      pref[1] = 1;
      auto getsum = [&](int l, int r) {
        return (pref[1 + r] - pref[l] + MOD) % MOD;
      };
      for (int i = 1; i <= n + 1; ++i) {
        pref[1 + i] = pref[i];
        for (auto [open, x, idx] : events[i]) {
          if (!(x & bit) && open) {
            right_start = max(right_start, open);
          }
        }
        if (ones[i]) continue;
        dp[i] = getsum(right_start, i - 1);
        pref[1 + i] = (pref[1 + i] + dp[i]) % MOD;
      }
      res = res * dp[n + 1] % MOD;
    }
    cout << res << "\n";
  }
}
