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;
const int N = 500 * 1000 + 13;
int n, k, m;
pair<pair<int, int>, int> q[N];
int ones[N];
int mx[N], sum[N];
int add(int first, int second) {
  first += second;
  if (first >= MOD) first -= MOD;
  if (first < 0) first += MOD;
  return first;
}
int calc(int t) {
  memset(ones, 0, sizeof(ones));
  memset(mx, -1, sizeof(mx));
  for (int i = 0; i < int(m); ++i) {
    int l = q[i].first.first, r = q[i].first.second;
    if (q[i].second & (1 << t)) {
      ones[l]++;
      ones[r + 1]--;
    } else {
      mx[r] = max(mx[r], l);
    }
  }
  int j = -1;
  for (int i = 0; i < int(n); ++i) {
    int cur = 0;
    if (!ones[i]) {
      cur = sum[i];
      if (j == -1)
        cur = add(cur, 1);
      else
        cur = add(cur, -sum[j]);
    }
    sum[i + 1] = add(sum[i], cur);
    ones[i + 1] += ones[i];
    j = max(j, mx[i]);
  }
  return add(sum[n], j != -1 ? -sum[j] : 1);
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 0; i < int(m); ++i) {
    scanf("%d%d%d", &q[i].first.first, &q[i].first.second, &q[i].second);
    --q[i].first.first;
    --q[i].first.second;
  }
  int ans = 1;
  for (int i = 0; i < int(k); ++i) ans = (ans * 1ll * calc(i)) % MOD;
  printf("%d\n", ans);
}
