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 long long MOD = 998244353;
const int MX = 5e5 + 233;
int read() {
  char k = getchar();
  int x = 0;
  while (k < '0' || k > '9') k = getchar();
  while (k >= '0' && k <= '9') x = x * 10 + k - '0', k = getchar();
  return x;
}
int n, k, m;
struct segment {
  int l, r, x;
  bool operator<(const segment &B) const {
    return r == B.r ? l > B.l : r < B.r;
  }
} s[MX];
long long ban[MX], S[MX], dp[MX];
long long solve(int bit) {
  for (int i = 1; i <= n + 1; ++i) ban[i] = 0;
  for (int i = 1; i <= m; ++i) {
    if ((s[i].x >> bit) & 1) {
      ban[s[i].l]++;
      ban[s[i].r + 1]--;
    }
  }
  int sl = 1, lim = 0;
  dp[0] = S[0] = 1;
  for (int i = 1; i <= n + 1; ++i) {
    ban[i] += ban[i - 1];
    while (sl <= m && s[sl].r < i) {
      if (((s[sl].x >> bit) & 1) == 0) {
        lim = max(lim, s[sl].l);
      }
      ++sl;
    }
    if (ban[i])
      dp[i] = 0;
    else if (lim)
      dp[i] = (S[i - 1] - S[lim - 1] + MOD) % MOD;
    else
      dp[i] = S[i - 1];
    S[i] = (S[i - 1] + dp[i]) % MOD;
  }
  return dp[n + 1];
}
int main() {
  n = read(), k = read(), m = read();
  for (int i = 1; i <= m; ++i) {
    s[i].l = read(), s[i].r = read(), s[i].x = read();
  }
  long long Ans = 1;
  sort(s + 1, s + 1 + m);
  for (int i = 0; i < k; ++i) {
    Ans = Ans * solve(i) % MOD;
  }
  cout << Ans << endl;
  return 0;
}
