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 l[500010], r[500010], x[500010], sum[500010], v[500010], qry[500010],
    dp[500010];
int main() {
  int n, k, m;
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; i++) scanf("%d%d%d", &l[i], &r[i], &x[i]);
  int ans = 1;
  for (int bit = 0; bit < k; bit++) {
    for (int i = 1; i <= n; i++) sum[i] = 0;
    for (int i = 1; i <= m; i++)
      if (x[i] & (1 << bit)) {
        sum[l[i]]++;
        sum[r[i] + 1]--;
      }
    int c = 0;
    for (int i = 1; i <= n; i++) {
      c += sum[i];
      if (c > 0)
        v[i] = 1;
      else
        v[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      sum[i] = sum[i - 1] + v[i];
      qry[i] = 0;
    }
    for (int i = 1; i <= m; i++)
      if (!(x[i] & (1 << bit))) {
        if (sum[r[i]] - sum[l[i] - 1] == r[i] - l[i] + 1) ans = 0;
        qry[r[i]] = max(qry[r[i]], l[i]);
      }
    dp[0] = 1;
    int s = 1, poz = 0;
    for (int i = 1; i <= n; i++) {
      if (v[i] == 1)
        dp[i] = 0;
      else
        dp[i] = s;
      s = (s + dp[i]) % mod;
      while (poz < qry[i]) {
        s -= dp[poz];
        s += mod;
        s %= mod;
        poz++;
      }
    }
    int sol = 0;
    for (int i = poz; i <= n; i++) sol = (sol + dp[i]) % mod;
    ans = (1LL * ans * sol) % mod;
  }
  printf("%d", ans);
  return 0;
}
