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;
const int mod = 998244353;
int n, k, m, c[maxn], pre[maxn], dp[maxn];
struct node {
  int l, r, x;
} a[maxn];
int read() {
  int res = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (!isdigit(ch));
  do {
    res = res * 10 + ch - '0';
    ch = getchar();
  } while (isdigit(ch));
  return res * f;
}
int calc(int ps) {
  for (int i = (1); i <= (n); ++i) c[i] = pre[i] = 0;
  for (int i = (1); i <= (m); ++i) {
    if (a[i].x & (1 << ps))
      c[a[i].l]++, c[a[i].r + 1]--;
    else
      pre[a[i].r] = max(pre[a[i].r], a[i].l);
  }
  for (int i = (1); i <= (n); ++i)
    c[i] += c[i - 1], pre[i] = max(pre[i], pre[i - 1]);
  int j = 0;
  for (int i = (1); i <= (n); ++i) {
    int num = 0;
    if (!c[i]) {
      if (!j)
        num = (dp[i] + 1) % mod;
      else
        num = (dp[i] - dp[j] + mod) % mod;
    }
    dp[i + 1] = (dp[i] + num) % mod;
    j = max(j, pre[i]);
  }
  if (!j)
    return (dp[n + 1] + 1) % mod;
  else
    return (dp[n + 1] - dp[j] + mod) % mod;
}
signed main() {
  n = read();
  k = read();
  m = read();
  for (int i = (1); i <= (m); ++i)
    a[i].l = read(), a[i].r = read(), a[i].x = read();
  long long ans = 1;
  for (int i = (0); i <= (k - 1); ++i) ans = ans * calc(i) % mod;
  printf("%lld\n", ans);
}
