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;
int Add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int Sub(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
int Mul(int x, int y) { return 1ll * x * y % mod; }
int n, k, m, ans;
int l[500005], r[500005], x[500005];
int Count[500005], Left[500005], dp[500005], psum[500005];
void Solve() {
  for (int i = 1; i <= n + 1; i++) {
    Count[i] = 0;
    Left[i] = 0;
    dp[i] = 0;
    psum[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    if (x[i] & 1) {
      Count[l[i]]++;
      Count[r[i] + 1]--;
    } else {
      Left[r[i]] = max(Left[r[i]], l[i]);
    }
  }
  for (int i = 1; i <= n; i++) {
    Count[i] += Count[i - 1];
    Left[i] = max(Left[i], Left[i - 1]);
    if (Count[i] == 0) {
      if (Left[i - 1] == 0) {
        dp[i] = Add(dp[i], 1);
      }
      dp[i] = Add(dp[i], Sub(psum[i - 1], psum[max(Left[i - 1] - 1, 0)]));
    }
    psum[i] = Add(psum[i - 1], dp[i]);
  }
  if (Left[n] == 0) {
    ans = Mul(ans, Add(psum[n], 1));
  } else {
    ans = Mul(ans, Sub(psum[n], psum[Left[n] - 1]));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ans = 1;
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) {
    cin >> l[i] >> r[i] >> x[i];
  }
  for (int i = 0; i < k; i++) {
    Solve();
    for (int j = 1; j <= m; j++) {
      x[j] >>= 1;
    }
  }
  cout << ans;
}
