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 MAXN = 5e5 + 5, Mod = 998244353;
char buf[1 << 12], *pp1 = buf, *pp2 = buf, nc;
int ny;
inline char gc() {
  return pp1 == pp2 &&
                 (pp2 = (pp1 = buf) + fread(buf, 1, 1 << 12, stdin), pp1 == pp2)
             ? EOF
             : *pp1++;
}
inline int read() {
  int x = 0;
  for (ny = 1; nc = gc(), (nc < 48 || nc > 57) && nc != EOF;)
    if (nc == 45) ny = -1;
  if (nc < 0) return nc;
  for (x = nc - 48; nc = gc(), 47 < nc && nc < 58 && nc != EOF;
       x = (x << 3) + (x << 1) + (nc ^ 48))
    ;
  return x * ny;
}
int n, k, m, ans = 1, s[MAXN], Mn[MAXN], l[MAXN], r[MAXN], c[MAXN], f[MAXN],
             q[MAXN], head, tail;
inline void add(int& x, int y) { x += y, x >= Mod ? x -= Mod : 0; }
inline void sub(int& x, int y) { x -= y, x < 0 ? x += Mod : 0; }
inline int Solve(int k) {
  memset(s, 0, sizeof(s));
  for (int i = 1; i <= n; i++) Mn[i] = n + 1;
  for (int i = 1; i <= m; i++) {
    if (c[i] >> k & 1)
      s[l[i]]++, s[r[i] + 1]--;
    else
      Mn[l[i]] = min(Mn[l[i]], r[i]);
  }
  for (int i = 1; i <= n; i++) s[i] += s[i - 1];
  memset(f, 0, sizeof(f)), f[n + 1] = 1, q[head = tail = 1] = n + 1;
  int sum = 1;
  for (int i = 1; i <= n; i++) {
    int tmp = 0;
    for (; q[tail] >= Mn[i] && head <= tail;
         add(tmp, f[q[tail]]), f[q[tail]] = 0, tail--)
      ;
    for (f[Mn[i]] = tmp, q[++tail] = Mn[i]; q[head] < i && head <= tail;
         sub(sum, f[q[head]]), head++)
      ;
    if (!s[i]) add(f[n + 1], sum), sum = 2 * sum % Mod;
    (q[tail] != n + 1 ? q[++tail] = n + 1 : 0);
  }
  return f[n + 1];
}
int main() {
  n = read(), k = read(), m = read();
  for (int i = 1; i <= m; i++) l[i] = read(), r[i] = read(), c[i] = read();
  for (int i = 0; i <= k - 1; i++) ans = 1ll * ans * Solve(i) % Mod;
  cout << ans << "\n";
  return 0;
}
