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 MOD = (119 << 23) + 1;
const int MX = 5e5;
int l[MX], r[MX], x[MX], one[MX + 2], dp[MX + 2], zer[MX + 2], ps[MX + 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k, m;
  cin >> n >> k >> m;
  for (int i = 0; i < m; ++i) {
    cin >> l[i] >> r[i] >> x[i];
  }
  long long ans = 1;
  for (int i = 1; i < (1 << k); i *= 2) {
    memset(ps, 0, sizeof ps);
    memset(dp, 0, sizeof dp);
    for (int j = 0; j < m; ++j) {
      if (x[j] & i) {
        ++one[l[j]];
        --one[r[j] + 1];
      } else {
        zer[r[j] + 1] = max(zer[r[j] + 1], l[j]);
      }
    }
    for (int i = 1; i <= n + 1; ++i) {
      one[i] += one[i - 1];
      zer[i] = max(zer[i], zer[i - 1]);
    }
    dp[0] = ps[0] = 1;
    for (int i = 1; i <= n + 1; ++i) {
      dp[i] = ps[i - 1];
      if (zer[i]) {
        ((dp[i] -= ps[zer[i] - 1]) += MOD) %= MOD;
        zer[i] = 0;
      }
      if (one[i]) {
        dp[i] = 0;
        one[i] = 0;
      }
      ps[i] = (dp[i] + ps[i - 1]) % MOD;
    }
    (ans *= dp[n + 1]) %= MOD;
  }
  cout << ans << '\n';
}
