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;
inline void chkmax(long long &a, long long b) {
  if (a < b) a = b;
}
inline void chkmin(long long &a, long long b) {
  if (a > b) a = b;
}
inline long long read() {
  long long a = 0, fh = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') fh = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    a = a * 10 + c - 48;
    c = getchar();
  }
  return a * fh;
}
long long n, m, k, sum[500005], f[500005], tag[500005], lim[500005];
;
vector<pair<long long, long long> > op[500005];
signed main() {
  n = read();
  k = read();
  m = read();
  for (long long i = 1; i <= m; ++i) {
    long long l = read(), r = read(), x = read();
    op[r].push_back(make_pair(l, x));
  }
  long long ans = 1;
  for (long long i = 0; i < k; ++i) {
    for (long long j = 1; j <= n + 1; ++j) sum[j] = f[j] = tag[j] = lim[j] = 0;
    sum[0] = f[0] = 1;
    for (long long j = 1; j <= n; ++j) {
      lim[j] = max(lim[j], lim[j - 1]);
      for (long long o = 0; o < op[j].size(); ++o) {
        long long l = op[j][o].first, r = j, x = op[j][o].second;
        if ((x >> i) & 1)
          tag[l]++, tag[r + 1]--;
        else
          lim[j + 1] = max(lim[j + 1], l);
      }
    }
    lim[n + 1] = max(lim[n + 1], lim[n]);
    for (long long j = 1; j <= n; ++j) tag[j] += tag[j - 1];
    for (long long j = 1; j <= n; ++j) {
      if (!tag[j])
        f[j] = (sum[j - 1] - (lim[j] ? sum[lim[j] - 1] : 0) + 998244353) %
               998244353;
      sum[j] = (sum[j - 1] + f[j]) % 998244353;
    }
    ans = ans * (sum[n] - (lim[n + 1] ? sum[lim[n + 1] - 1] : 0) + 998244353) %
          998244353;
  }
  printf("%lld\n", ans);
  return 0;
}
