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;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
inline int add(int x, int y) {
  return x + y >= 998244353 ? x + y - 998244353 : x + y;
}
inline int sub(int x, int y) { return x - y < 0 ? x - y + 998244353 : x - y; }
inline int mul(int x, int y) { return 1LL * x * y % 998244353; }
int n, k, m, ans;
int l[500005], r[500005], x[500005], f[500005], lim[500005], a[500005];
int solve(int u) {
  memset(f, 0, sizeof(f));
  memset(a, 0, sizeof(a));
  memset(lim, 0, sizeof(lim));
  for (int i = 1; i <= m; i++) {
    if (x[i] & (1 << u))
      a[l[i]]++, a[r[i] + 1]--;
    else
      lim[r[i]] = max(lim[r[i]], l[i]);
  }
  for (int i = 1; i <= n + 1; i++) a[i] += a[i - 1];
  f[0] = 1;
  int l = 0, sum = 1;
  for (int i = 1; i <= n + 1; i++) {
    while (l < lim[i - 1]) sum = sub(sum, f[l++]);
    if (a[i]) continue;
    f[i] = sum;
    sum = add(sum, f[i]);
  }
  return f[n + 1];
}
signed main() {
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) l[i] = read(), r[i] = read(), x[i] = read();
  ans = 1;
  for (int i = 0; i < k; i++) ans = mul(ans, solve(i));
  cout << ans << endl;
  return 0;
}
