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;
struct condition {
  int l, r, x;
  condition(int _l, int _r, int _x) : l(_l), r(_r), x(_x) {}
};
vector<condition> conds;
int n, k, m;
int fix[500005];
vector<condition> st;
int range[500005];
long long any[500005], zero[500005], zs[500005];
long long sumz(int i, int j) {
  if (i > j) return 0ll;
  return (zs[j] - zs[i - 1] + 998244353ll) % 998244353ll;
}
long long ways(int idx) {
  memset(fix, 0, sizeof(fix));
  st.clear();
  for (auto c : conds) {
    if (c.x & (1 << idx)) {
      fix[c.l]++;
      fix[c.r + 1]--;
    } else {
      if (!st.empty() && st.back().l >= c.l)
        continue;
      else
        st.push_back(c);
    }
  }
  memset(range, -1, sizeof(range));
  for (int i = 2; i <= n; i++) fix[i] += fix[i - 1];
  for (int i = 1; i <= n; i++) fix[i] = !!fix[i];
  for (auto c : st) {
    range[c.r] = c.l;
  }
  any[0] = 1ll;
  zero[0] = 1ll;
  zs[0] = 0ll;
  for (int i = 1; i <= n; i++) {
    if (range[i] == -1) {
      if (fix[i]) {
        zero[i] = 0ll;
        any[i] = any[i - 1];
      } else {
        zero[i] = any[i - 1];
        any[i] = any[i - 1] * 2ll;
      }
    } else {
      if (fix[i]) {
        zero[i] = 0ll;
        any[i] = sumz(range[i], i - 1);
      } else {
        zero[i] = any[i - 1];
        any[i] = any[i - 1] + sumz(range[i], i - 1);
      }
    }
    zero[i] %= 998244353ll;
    any[i] %= 998244353ll;
    zs[i] = (zs[i - 1] + zero[i]) % 998244353ll;
  }
  return any[n];
}
bool cmp(condition &x, condition &y) {
  if (x.r == y.r) return x.l > y.l;
  return x.r < y.r;
}
int main() {
  scanf("%d%d%d", &n, &k, &m);
  int l, r, x;
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &l, &r, &x);
    conds.push_back(condition(l, r, x));
  }
  sort((conds).begin(), (conds).end(), cmp);
  long long ans = 1ll;
  for (int i = 0; i < k; i++) {
    long long r = ways(i);
    ans = (ans * r) % 998244353ll;
  }
  printf("%lld\n", ans);
}
