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;
const int INF = 1e9, MOD = 998244353;
int n, m;
vector<int> dp;
vector<bool> ones;
int64_t solve(vector<pair<int, int> >& yes, vector<pair<int, int> >& no) {
  sort(yes.begin(), yes.end());
  for (int i = 1; i < n + 1; i++) ones[i] = 0;
  int last = -1;
  for (auto seg : yes) {
    if (last < seg.first) last = seg.first;
    for (; last <= seg.second; last++) ones[last] = 1;
  }
  for (auto& p : no) swap(p.first, p.second);
  sort(no.begin(), no.end());
  dp[0] = 1;
  int maxs = 0, sum = 1, cur = 0;
  for (int i = 1; i < n + 2; i++) {
    while (cur < no.size() && no[cur].first < i) {
      while (maxs < no[cur].second) {
        sum = (sum - dp[maxs++] + MOD) % MOD;
      }
      cur++;
    }
    if (ones[i]) {
      dp[i] = 0;
      continue;
    }
    dp[i] = sum;
    sum = (sum + dp[i]) % MOD;
  }
  return dp[n + 1];
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  int k;
  cin >> n >> k >> m;
  vector<vector<pair<int, int> > > yes(k), no(k);
  dp = vector<int>(n + 2);
  ones = vector<bool>(n + 1);
  for (int i = 0; i < m; i++) {
    int l, r, first;
    cin >> l >> r >> first;
    for (int j = 0; j < k; j++)
      if (first & (1 << j))
        yes[j].push_back({l, r});
      else
        no[j].push_back({l, r});
  }
  int64_t ans = 1;
  for (int i = 0; i < k; i++) ans = (ans * solve(yes[i], no[i])) % MOD;
  cout << ans << endl;
  return 0;
}
