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 N = 5e5 + 1, mod = 998244353;
int n, k, m, valq[N];
int cnt[30][N], prv[30][N], dp[30][N];
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
int binpow(int a, int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  int ret = binpow(a, b / 2);
  if (b % 2 == 0) return 1LL * ret * ret % mod;
  return 1LL * ret * ret % mod * a % mod;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) {
    int l, r;
    cin >> l >> r >> valq[i];
    for (int j = 0; j < k; j++) {
      if (valq[i] & (1 << j)) {
        cnt[j][l]++, cnt[j][r + 1]--;
      } else {
        prv[j][r] = max(prv[j][r], l);
      }
    }
  }
  int ans = 1;
  for (int i = 0; i < k; i++) {
    dp[i][0] = 1;
    int sum = 0, pos = 0;
    for (int j = 1; j <= n; j++) {
      sum += cnt[i][j];
      if (sum > 0)
        dp[i][j] = dp[i][j - 1];
      else
        dp[i][j] =
            sub(2LL * dp[i][j - 1] % mod, (pos - 1 >= 0 ? dp[i][pos - 1] : 0));
      if (prv[i][j] != 0) pos = max(pos, prv[i][j]);
    }
    ans = 1LL * ans * sub(dp[i][n], (pos - 1 >= 0 ? dp[i][pos - 1] : 0)) % mod;
  }
  cout << ans;
}
