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 = 5e5 + 2;
const int mod = 998244353;
void inc(int& x, const int& y) { (x += y) >= mod ? x -= mod : 0; }
void dec(int& x, const int& y) { x < y ? x += mod - y : x -= y; }
void cmax(int& x, const int& y) { x = x > y ? x : y; }
int n, k, m;
int l[N], r[N], x[N], pos[N], a[N], f[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) cin >> l[i] >> r[i] >> x[i];
  int ans = 1;
  for (int p = 0; p <= k - 1; p++) {
    for (int i = 1; i <= n + 1; i++) pos[i] = a[i] = 0;
    for (int i = 1; i <= m; i++)
      if (x[i] >> p & 1)
        ++a[l[i]], --a[r[i] + 1];
      else
        cmax(pos[r[i] + 1], l[i]);
    for (int i = 2; i <= n + 1; i++) a[i] += a[i - 1], cmax(pos[i], pos[i - 1]);
    f[0] = 1;
    int sum = 1, l = 0;
    for (int i = 1; i <= n + 1; i++) {
      while (l < pos[i]) dec(sum, f[l]), f[l++] = 0;
      f[i] = a[i] ? 0 : sum, inc(sum, f[i]);
    }
    ans = 1ll * ans * f[n + 1] % mod;
  }
  cout << ans << '\n';
  return 0;
}
