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 MAX = 500010;
const int MOD = 998244353;
inline int add(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
inline int sub(int a, int b) { return a - b < 0 ? a - b + MOD : a - b; }
inline int mul(int a, int b) { return (a * 1LL * b) % MOD; }
int n, m, k;
int L[MAX], R[MAX], X[MAX];
int lft[MAX];
int dp[MAX];
int cum[MAX];
int fix[MAX];
int solve(int b) {
  memset(fix, 0, sizeof(fix));
  memset(lft, 0, sizeof(lft));
  for (int i = 1; i <= m; i++) {
    if (X[i] & (1 << b)) {
      fix[L[i]]++;
      fix[R[i] + 1]--;
    } else {
      lft[R[i]] = max(lft[R[i]], L[i]);
    }
  }
  for (int i = 1; i <= n; i++) fix[i] += fix[i - 1];
  dp[0] = 1;
  cum[0] = 1;
  int carry = -1;
  for (int i = 1; i <= n; i++) {
    if (carry == 0) {
      if (fix[i])
        dp[i] = 0;
      else
        dp[i] = cum[i - 1];
    } else {
      if (fix[i])
        dp[i] = 0;
      else
        dp[i] = sub(cum[i - 1], cum[carry - 1]);
    }
    carry = max(carry, lft[i]);
    cum[i] = add(cum[i - 1], dp[i]);
  }
  if (carry == 0)
    return sub(cum[n], cum[carry - 1]);
  else
    return sub(cum[n], cum[carry - 1]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k >> m;
  for (int i = 1; i <= m; i++) cin >> L[i] >> R[i] >> X[i];
  int ans = 1;
  for (int i = 0; i < k; i++) {
    ans = mul(ans, solve(i));
  }
  cout << ans << endl;
  return 0;
}
