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 long long N = 5e5 + 5;
long long n, k, m;
long long pos[N];
struct range {
  long long l, r, w;
  bool operator<(const range& A) const { return l < A.l; }
} p[N];
long long v[N];
long long sum[N];
long long solve(long long k) {
  memset(v, 0, sizeof(v));
  memset(pos, 0, sizeof(pos));
  for (long long i = 1; i <= m; i++)
    if (p[i].w & (1 << k - 1))
      v[p[i].l]++, v[p[i].r + 1]--;
    else
      pos[p[i].r] = max(pos[p[i].r], p[i].l);
  for (long long i = 2; i <= n + 1; i++)
    v[i] += v[i - 1], pos[i] = max(pos[i], pos[i - 1]);
  long long now = 0;
  sum[0] = 1;
  for (long long i = 1; i <= n + 1; i++) {
    sum[i] = sum[i - 1];
    if (v[i]) continue;
    now = max(now, pos[i - 1]);
    (sum[i] += sum[i - 1] - ((now - 1) < 0 ? 0 : sum[now - 1])) %= mod;
  }
  return (sum[n + 1] - sum[n] + mod) % mod;
}
signed main() {
  scanf("%lld %lld %lld", &n, &k, &m);
  for (long long i = 1; i <= m; i++)
    scanf("%lld %lld %lld", &p[i].l, &p[i].r, &p[i].w);
  long long ans = 1;
  for (long long i = 1; i <= k; i++) ans = 1ll * ans * solve(i) % mod;
  printf("%lld", (ans + mod) % mod);
  return 0;
}
