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;
struct que {
  int l, r, x;
} arr[500005];
int tag[500005], state[500005];
int L[500005];
int dp[500005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k, m;
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) cin >> arr[i].l >> arr[i].r >> arr[i].x;
  int ans = 1;
  for (int bit = 0; bit < k; bit++) {
    int b = (1 << bit);
    memset(tag, 0, sizeof tag);
    memset(L, 0, sizeof L);
    for (int i = 1; i <= m; i++)
      if ((arr[i].x & b) != 0)
        tag[arr[i].l] = max(tag[arr[i].l], arr[i].r);
      else
        L[arr[i].r] = max(L[arr[i].r], arr[i].l);
    memset(state, 0, sizeof state);
    for (int i = 1, r = 0; i <= n; i++) {
      r = max(r, tag[i]);
      state[i] = (r >= i ? 1 : 0);
    }
    memset(dp, 0, sizeof dp);
    dp[0] = 1;
    int now = 1, l = 0;
    for (int i = 1; i <= n; i++) {
      if (state[i] == 0) dp[i] = (((long long)now) % 998244353);
      for (; l < L[i]; l++)
        now = (((long long)now - dp[l] + 998244353) % 998244353);
      now = (((long long)now + dp[i]) % 998244353);
    }
    ans = (((long long)ans * now) % 998244353);
  }
  cout << ans << endl;
  return 0;
}
