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, K = 30, mod = 998244353;
struct Node {
  int l, r, x;
};
int n, m, k;
Node a[maxn];
int p[maxn], ones[maxn];
int sum[maxn], dp[maxn];
int calc(int t) {
  memset(p, 0, sizeof(p));
  memset(ones, 0, sizeof(ones));
  for (int i = 0; i < m; i++) {
    int l = a[i].l, r = a[i].r, x = a[i].x;
    if (x >> t & 1) {
      ones[l]++;
      ones[r + 1]--;
    } else {
      p[r] = max(p[r], l);
    }
  }
  int j = 0;
  for (int i = 1; i <= n + 1; i++) {
    ones[i] += ones[i - 1];
    dp[i] = 0;
    if (!ones[i]) {
      if (!j) {
        dp[i] = (sum[i - 1] + 1) % mod;
      } else {
        dp[i] = ((sum[i - 1] - sum[j - 1]) % mod + mod) % mod;
      }
    }
    sum[i] = (sum[i - 1] + dp[i]) % mod;
    j = max(j, p[i]);
  }
  return dp[n + 1];
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].x);
  }
  int res = 1;
  for (int i = 0; i < k; i++) {
    res = (res * 1ll * calc(i)) % mod;
  }
  printf("%d\n", res);
  return 0;
}
