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;
char _buf[100000], *_op(_buf), *_ed(_buf);
inline int getint() {
  int _s = 0, _f = 1;
  char _ch = (_op == _ed && (_ed = (_op = _buf) + fread(_buf, 1, 100000, stdin),
                             _op == _ed)
                  ? EOF
                  : *_op++);
  while (!('0' <= _ch && _ch <= '9'))
    (_ch == '-') && (_f = -1),
        _ch =
            (_op == _ed && (_ed = (_op = _buf) + fread(_buf, 1, 100000, stdin),
                            _op == _ed)
                 ? EOF
                 : *_op++);
  while (('0' <= _ch && _ch <= '9'))
    _s = _s * 10 + _ch - 48,
    _ch = (_op == _ed && (_ed = (_op = _buf) + fread(_buf, 1, 100000, stdin),
                          _op == _ed)
               ? EOF
               : *_op++);
  return _s * _f;
}
inline int min(const int& x, const int& y) { return (x - y >> 31) ? x : y; }
inline int max(const int& x, const int& y) { return (y - x >> 31) ? x : y; }
const int N = 5e5 + 2, mod = 998244353;
int n, m, k, ans = 1;
int l[N], r[N], a[N], f[N], lim[N], d[N];
inline void check(int& x) { x -= mod, x += x >> 31 & mod; }
inline void input() {
  n = getint(), k = getint(), m = getint();
  for (int i = 1; i <= m; i++)
    l[i] = getint(), r[i] = getint(), a[i] = getint();
}
inline void init() {}
inline void test(int p) {
  int cur = 0, sum = 0;
  for (int i = 0; i <= n + 1; i++) lim[i] = 0, d[i] = 0, f[i] = 0;
  for (int i = 1; i <= m; i++)
    if (a[i] >> p & 1)
      d[l[i]]++, d[r[i] + 1]--;
    else
      lim[r[i]] = max(lim[r[i]], l[i]);
  f[0] = 1, sum = 1;
  for (int i = 1; i <= n; i++) {
    if (!(d[i] += d[i - 1])) f[i] = sum, check(sum <<= 1);
    while (cur < lim[i]) check(sum += mod - f[cur++]);
  }
  ans = 1ll * ans * sum % mod;
}
inline void work() {
  for (int i = 0; i <= k - 1; i++) test(i);
  cout << ans;
}
int main() {
  input();
  init();
  work();
  return 0;
}
