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 = 998244353;
int n, k, t;
int f[500005], mx[500005], sum[500005];
struct query {
  int first, second, c;
} q[500001];
inline int add(int first, int second) {
  first += second;
  if (first >= mod) first -= mod;
  if (first < 0) first += mod;
  return first;
}
inline long long compute(int b) {
  memset(f, 0, sizeof(f));
  memset(mx, -1, sizeof(mx));
  for (int i = 1; i <= t; ++i) {
    int l = q[i].first, r = q[i].second;
    if (q[i].c & (1 << b)) {
      ++f[l];
      --f[r + 1];
    } else
      mx[r] = max(mx[r], l);
  }
  int j = -1;
  for (int i = 1; i <= n; ++i) {
    int aux = 0;
    if (!f[i]) {
      aux = sum[i];
      if (j == -1)
        aux = add(aux, 1);
      else
        aux = add(aux, -sum[j]);
    }
    sum[i + 1] = add(sum[i], aux);
    f[i + 1] += f[i];
    j = max(j, mx[i]);
  }
  return add(sum[n + 1], (j != -1 ? -sum[j] : 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> n >> k >> t;
  for (int first, j = 1; j <= t; ++j)
    cin >> q[j].first >> q[j].second >> q[j].c;
  long long rez = 1;
  for (int i = 0; i < k; ++i) rez = (1LL * rez * compute(i)) % mod;
  return cout << rez, 0;
}
