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;
int n, m, k;
pair<pair<int, int>, int> req[500005];
int ops[500005];
long long dp[500005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> m;
  for (int i = 0; i < (signed)(m); i++) {
    cin >> req[i].first.second >> req[i].first.first >> req[i].second;
  }
  sort(req, req + m);
  long long ans = 1;
  for (int bt = 0; bt < (signed)(k); bt++) {
    memset(ops, 0, sizeof ops);
    for (int i = 0; i < (signed)(m); i++) {
      if (req[i].second & (1 << bt)) {
        ops[req[i].first.second]++;
        ops[req[i].first.first + 1]--;
      }
    }
    memset(dp, 0, sizeof dp);
    int lb = 0;
    int cur = 0;
    int cc = 0;
    long long csm = 1;
    dp[0] = 1;
    for (int i = (1); i <= (signed)(n); i++) {
      cc += ops[i];
      if (cc == 0) {
        dp[i] = csm;
        csm = (csm + dp[i]) % 998244353LL;
      }
      while (cur < m && req[cur].first.first == i) {
        if (!((1 << bt) & req[cur].second)) {
          int olb = lb;
          lb = max(lb, req[cur].first.second);
          for (int i = (olb); i <= (signed)(lb - 1); i++) {
            csm = (csm + 998244353LL - dp[i]) % 998244353LL;
          }
        }
        ++cur;
      }
    }
    long long tot = csm;
    ans = (ans * tot) % 998244353LL;
  }
  cout << ans << "\n";
  return 0;
}
