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 N = 500100;
const int md = 998244353;
int ans = 1, n, k, m, pl[N], mx[N], pf[N], f[N], x[N], l[N], r[N];
bool mrk[N];
int mult(int x, int y) { return (1ll * x * y) % md; }
int sum(int x, int y) {
  x += y;
  if (x >= md) x -= md;
  return x;
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) x += md;
  return x;
}
int calc(int bt) {
  fill(pl, pl + n + 1, 0);
  fill(mx, mx + n + 1, 0);
  int mx_l = 0;
  for (int i = 0; i < m; i++)
    if (x[i] & (1 << bt)) {
      pl[l[i]]++;
      pl[r[i] + 1]--;
    } else {
      mx_l = max(mx_l, l[i]);
      mx[r[i] + 1] = max(mx[r[i] + 1], l[i]);
    }
  int sm = 0;
  for (int i = 1; i <= n; i++) {
    sm += pl[i];
    mrk[i] = bool(sm > 0);
  }
  for (int i = 1; i <= n; i++) mx[i] = max(mx[i], mx[i - 1]);
  int res = 0;
  f[0] = 1;
  if (0 >= mx_l) res = 1;
  for (int i = 1; i <= n; i++)
    if (mrk[i]) {
      f[i] = f[i - 1];
    } else {
      f[i] = sub(f[i - 1], (mx[i] == 0 ? 0 : f[mx[i] - 1]));
      if (mx_l <= i) res = sum(res, f[i]);
      f[i] = sum(f[i], f[i - 1]);
    }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> m;
  for (int i = 0; i < m; i++) {
    cin >> l[i] >> r[i] >> x[i];
  }
  for (int bt = 0; bt < k; bt++) ans = mult(ans, calc(bt));
  cout << ans;
  return 0;
}
