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 inf = 1e9;
const int Mod = 998244353;
const int Nmax = 5e5 + 5;
int l[Nmax], r[Nmax], res[Nmax];
int start1[Nmax], end0[Nmax], need[Nmax], dp[Nmax], s[Nmax];
int n, m, k;
void inm_to(int &x, int y) { x = (long long)x * y % Mod; }
int subs(int x, int y) {
  x -= y;
  if (x < 0) return x + Mod;
  return x;
}
int add(int x, int y) {
  x += y;
  if (x >= Mod) return x - Mod;
  return x;
}
int solve(int b) {
  int i;
  memset(start1, 0, sizeof(start1));
  memset(end0, 0, sizeof(end0));
  memset(need, 0, sizeof(need));
  for (i = 1; i <= m; ++i)
    if (res[i] & (1 << b))
      start1[l[i]] = max(start1[l[i]], r[i] - l[i] + 1);
    else
      end0[r[i]] = max(end0[r[i]], l[i]);
  for (i = 1; i <= n; ++i)
    if (start1[i] > 0) {
      need[i] = 1;
      start1[i + 1] = max(start1[i + 1], start1[i] - 1);
    }
  dp[0] = 1;
  s[0] = 1;
  int x = 0;
  for (i = 1; i <= n; ++i) {
    x = max(x, end0[i - 1]);
    if (need[i])
      dp[i] = 0;
    else
      dp[i] = subs(s[i - 1], (x ? s[x - 1] : 0));
    s[i] = add(s[i - 1], dp[i]);
  }
  x = max(x, end0[n]);
  return subs(s[n], (x ? s[x - 1] : 0));
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> m;
  int i;
  for (i = 1; i <= m; ++i) cin >> l[i] >> r[i] >> res[i];
  int ans = 1;
  for (i = 0; i < k; ++i) inm_to(ans, solve(i));
  cout << ans << '\n';
  return 0;
}
