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, Maxn = 500005;
inline void read(int& x) {
  static char c;
  while (!isdigit(c = getchar()))
    ;
  x = c ^ 48;
  while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
}
inline int Max(int a, int b) { return a > b ? a : b; }
inline int Add(int a, int b) { return (a += b) > Mod ? a - Mod : a; }
inline int Del(int a, int b) { return (a -= b) < 0 ? a + Mod : a; }
int n, k, m;
struct Opt {
  int l, r, x;
} opt[Maxn];
int dp[Maxn], del[Maxn], l[Maxn], pi, s, pos;
inline int Solve(int nk) {
  s = 1, pos = 0, memset(del, 0, sizeof del), memset(l, 0, sizeof l),
  memset(dp, 0, sizeof dp);
  for (register int i = 1; i <= m; ++i)
    if (opt[i].x & (1 << nk))
      ++del[opt[i].l], --del[opt[i].r + 1];
    else
      l[opt[i].r] = Max(l[opt[i].r], opt[i].l);
  for (register int i = 1; i <= n + 1; ++i)
    l[i] = Max(l[i], l[i - 1]), del[i] += del[i - 1];
  dp[0] = 1;
  for (register int i = 1; i <= n + 1; ++i) {
    if (del[i]) {
      dp[i] = 0;
      continue;
    }
    while (pos < l[i - 1]) s = Del(s, dp[pos++]);
    dp[i] = s, s = Add(s, dp[i]);
  }
  return dp[n + 1];
}
inline void Work() {
  read(n), read(k), read(m);
  for (register int i = 1; i <= m; ++i)
    read(opt[i].l), read(opt[i].r), read(opt[i].x);
  int Ans = 1;
  for (register int i = 0; i < k; ++i) Ans = 1ll * Ans * Solve(i) % Mod;
  printf("%d\n", Ans);
  getchar(), getchar();
}
int main() { Work(); }
