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;
using pii = pair<int, int>;
const int N = 5e5 + 5, M = 998244353;
int n, k, m, ls[N], rs[N], xs[N];
int solve(int bit) {
  vector<int> dp(n + 2), a(n + 2), asum(n + 2), dpsum(n + 2), f(n + 2);
  vector<pii> ranges;
  for (int i = 1; i <= m; ++i) {
    if (xs[i] & 1 << bit) {
      ++a[ls[i]], --a[rs[i] + 1];
    } else {
      ranges.emplace_back(ls[i], rs[i]);
    }
  }
  sort(ranges.begin(), ranges.end(),
       [](pii x, pii y) { return x.second < y.second; });
  for (int i = 1, idx = 0; i <= n + 1; ++i) {
    f[i] = f[i - 1];
    while (idx < ranges.size() && ranges[idx].second < i) {
      f[i] = max(f[i], ranges[idx++].first);
    }
  }
  partial_sum(a.begin() + 1, a.begin() + 1 + n, asum.begin() + 1);
  dp[0] = dpsum[0] = 1;
  for (int i = 1; i <= n + 1; ++i) {
    if (asum[i]) {
      dp[i] = 0;
    } else {
      dp[i] = (dpsum[i - 1] - (f[i] ? dpsum[f[i] - 1] : 0) + M) % M;
    }
    dpsum[i] = (dpsum[i - 1] + dp[i]) % M;
  }
  return dp[n + 1];
}
int main() {
  cin >> n >> k >> m;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d%d", &ls[i], &rs[i], &xs[i]);
  }
  int ans = 1;
  for (int i = 0; i < k; ++i) ans = 1ll * ans * solve(i) % M;
  cout << ans << endl;
}
