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;
int n, m, K, set_zero;
int l[maxn], r[maxn], x[maxn][50], c[maxn], key[maxn];
long long dp[maxn], sum[maxn];
inline int lower_bit(int x) { return x & (-x); }
void insert(int k, int x) {
  for (int i = k; i <= n; i += lower_bit(i)) c[i] += x;
}
int ask(int x) {
  int ss = 0;
  while (x) {
    ss += c[x];
    x -= lower_bit(x);
  }
  return ss;
}
int main() {
  int y;
  scanf("%d%d%d", &n, &K, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &l[i], &r[i], &y);
    for (int k = 0; k < K; k++) {
      x[i][k] = y % 2;
      y /= 2;
    }
  }
  long long ans = 1;
  for (int k = 0; k < K; k++) {
    for (int i = 1; i <= n + 1; i++)
      c[i] = 0, key[i] = 0, dp[i] = 0, sum[i] = 0;
    for (int i = 1; i <= m; i++) {
      if (x[i][k])
        insert(l[i], 1), insert(r[i] + 1, -1);
      else
        key[r[i]] = max(key[r[i]], l[i]);
    }
    dp[0] = 1;
    set_zero = -1;
    sum[0] = 1;
    for (int i = 1; i <= n; i++) {
      if (ask(i))
        dp[i] = 0;
      else
        dp[i] = (sum[i - 1] - (set_zero >= 0 ? sum[set_zero] : 0) + mod) % mod;
      sum[i] = (sum[i - 1] + dp[i]) % mod;
      if (key[i]) set_zero = max(set_zero, key[i] - 1);
    }
    ans = ans * ((sum[n] - (set_zero >= 0 ? sum[set_zero] : 0) + mod) % mod) %
          mod;
  }
  cout << ans;
  return 0;
}
