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 long long MOD = 998244353;
inline long long add(long long a, long long b) {
  long long ret = a + b;
  while (ret >= MOD) {
    ret -= MOD;
  }
  while (ret < 0) {
    ret += MOD;
  }
  return ret;
}
inline long long mult(long long a, long long b) { return (a * b) % MOD; }
vector<pair<pair<long long, long long>, long long>> edge;
long long n, k, m;
long long solve(long long biti) {
  vector<long long> seg(n + 1, 0), seg_fill(n + 1, 0);
  for (auto it : edge) {
    if (it.second & (1 << biti)) {
      seg_fill[it.first.first]++;
      seg_fill[it.first.second + 1]--;
    }
  }
  long long count = 0;
  for (int i = 0; i < n + 1; i++) {
    count += seg_fill[i];
    if (count > 0) {
      seg[i] = 1;
    }
  }
  vector<long long> pre(n + 1, 0);
  for (auto it : edge) {
    if (!(it.second & (1 << biti))) {
      pre[it.first.second + 1] =
          max(pre[it.first.second + 1], it.first.first + 1);
    }
  }
  for (int i = 1; i <= n; i++) {
    pre[i] = max(pre[i], pre[i - 1]);
  }
  vector<long long> dp(n + 2, 0);
  dp[0] = 1;
  vector<long long> pre_dp(n + 2, 0);
  pre_dp[0] = 0;
  for (int i = 1; i <= n + 1; i++) {
    pre_dp[i] = add(pre_dp[i - 1], dp[i - 1]);
    if (!seg[i - 1]) {
      dp[i] = add(pre_dp[i], -pre_dp[pre[i - 1]]);
    }
  }
  return dp[n + 1];
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k >> m;
  edge.resize(m);
  for (auto &it : edge) {
    cin >> it.first.first >> it.first.second >> it.second;
    it.first.first--, it.first.second--;
  }
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    ans = mult(ans, solve(i));
  }
  cout << ans << "\n";
}
