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>
#pragma GCC optimize("O3")
const double PI = acos(-1);
template <class A, class B>
std::ostream& operator<<(std::ostream& st, const std::pair<A, B> p) {
  st << "(" << p.first << ", " << p.second << ")";
  return st;
}
using namespace std;
const int mod = 998244353;
int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  if (a < 0) a += mod;
  return a;
}
int mul(int a, int b) { return (a * (long long)b) % mod; }
const int MAX = 500 * 1000 + 10;
int n, K, m;
int L[MAX], R[MAX], X[MAX];
int cnt[MAX];
int val[MAX];
int tail[MAX];
int p2[MAX];
int dp[MAX], sum[MAX];
int prv[MAX], nxt[MAX];
int solve(int b) {
  static std::vector<std::pair<int, int> > V;
  for (int i = 0; i < (n + 1); i++) cnt[i] = 0;
  V.clear();
  for (int i = 0; i < (m); i++) {
    int x = (X[i] >> b) & 1;
    if (x) {
      cnt[L[i]]++;
      cnt[R[i] + 1]--;
    } else {
      V.emplace_back(L[i], R[i]);
    }
  }
  for (int i = 1; i <= n; i++) {
    cnt[i] += cnt[i - 1];
    val[i] = (cnt[i] > 0) + val[i - 1];
  }
  for (int i = 1, v = -1; i <= n; i++) {
    if (!cnt[i]) v = i;
    prv[i] = v;
  }
  for (int i = n, v = -1; i >= 1; i--) {
    if (!cnt[i]) v = i;
    nxt[i] = v;
  }
  for (int i = 1; i <= n; i++) tail[i] = -1;
  for (auto p : V) {
    int l = p.first, r = p.second;
    int ctr = val[r] - val[l - 1];
    if (ctr == r - l + 1) return 0;
    l = nxt[l];
    r = prv[r];
    l = l - val[l];
    r = r - val[r];
    tail[r] = max(tail[r], l);
  }
  int m = n - val[n];
  if (!m) return 1;
  dp[0] = 1;
  sum[0] = 1;
  int L = 0;
  for (int i = 1; i <= m; i++) {
    dp[i] = add(sum[i - 1], L ? -sum[L - 1] : 0);
    sum[i] = add(dp[i], sum[i - 1]);
    L = max(L, tail[i]);
  }
  int ret = add(sum[m], L ? -sum[L - 1] : 0);
  return ret;
}
int main() {
  p2[0] = 1;
  for (int i = 1; i < MAX; i++) p2[i] = mul(2, p2[i - 1]);
  scanf("%d %d %d", &n, &K, &m);
  for (int i = 0; i < (m); i++) scanf("%d %d %d", L + i, R + i, X + i);
  int ans = 1;
  for (int b = 0; b < (K); b++) {
    ans = mul(ans, solve(b));
    if (!ans) break;
  }
  cout << ans << endl;
  return 0;
}
