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>
const int maxn = 500005, maxk = 31, mod = 998244353;
int n, k, m, ans;
int cnt[maxk][maxn], maxx[maxk][maxn], f[maxn];
inline int max(int a, int b) { return a > b ? a : b; }
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; i++) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    for (int j = 0; j < k; j++) {
      if ((x >> j) & 1)
        cnt[j][l]++, cnt[j][r + 1]--;
      else
        maxx[j][r + 1] = max(maxx[j][r + 1], l);
    }
  }
  for (int i = 0; i < k; i++)
    for (int j = 1; j <= n + 1; j++) {
      cnt[i][j] += cnt[i][j - 1];
      maxx[i][j] = max(maxx[i][j], maxx[i][j - 1]);
    }
  ans = 1;
  for (int i = 0; i < k; i++) {
    int pos = 0, sum = 0;
    for (int j = 0; j <= n + 1; j++) f[j] = 0;
    f[0] = sum = 1;
    for (int j = 1; j <= n + 1; j++) {
      while (pos < maxx[i][j]) sum = (sum - f[pos] + mod) % mod, pos++;
      if (cnt[i][j] == 0)
        f[j] = sum;
      else
        f[j] = 0;
      sum = (sum + f[j]) % mod;
    }
    ans = 1ll * ans * f[n + 1] % mod;
  }
  printf("%d\n", ans);
  return 0;
}
