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;
int Mod(int x) { return x >= mod ? x - mod : x; }
long long read() {
  long long x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
  return x * f;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
const int N = 500010;
int n, k, m, l[N], r[N], x[N], c[N], lmx[N];
long long ans = 1, dp[N];
long long solve(int s) {
  for (int i = 1; i <= n; i++) lmx[i] = c[i] = dp[i] = 0;
  for (int i = 1; i <= m; i++)
    if (x[i] >> s & 1)
      c[l[i]]++, c[r[i] + 1]--;
    else
      lmx[r[i]] = max(lmx[r[i]], l[i]);
  for (int i = 1; i <= n + 1; i++)
    lmx[i] = max(lmx[i], lmx[i - 1]), c[i] += c[i - 1];
  long long res = 1, j = 0;
  dp[0] = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (c[i] > 0) {
      dp[i] = 0;
      continue;
    }
    while (j < lmx[i - 1]) res = Mod(res - dp[j++] + mod);
    dp[i] = res;
    res = (res + dp[i]) % mod;
  }
  return dp[n + 1];
}
int main() {
  n = read(), k = read(), m = read();
  for (int i = 1; i <= m; i++) l[i] = read(), r[i] = read(), x[i] = read();
  for (int i = 0; i < k; i++) ans = ans * solve(i) % mod;
  print((ans % mod + mod) % mod), putchar('\n');
  return 0;
}
