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 NMAX = 5e5 + 5, KMAX = 35;
const int MOD = 998244353;
int N, K, M;
vector<pair<int, int> > Restrictions[KMAX];
int Mars[KMAX][NMAX];
int dp[NMAX];
int sp[NMAX];
static inline void Read() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> K >> M;
  for (int i = 1; i <= M; ++i) {
    int l = 0, r = 0, x = 0;
    cin >> l >> r >> x;
    for (int b = 0; b < K; ++b) {
      if (x & (1 << b))
        Mars[b][l] += 1, Mars[b][r + 1] -= 1;
      else
        Restrictions[b].push_back({l, r});
    }
  }
  return;
}
static inline void Precalculation() {
  for (int b = 0; b < K; ++b)
    for (int i = 2; i <= N; ++i) Mars[b][i] += Mars[b][i - 1];
  return;
}
static inline void Reset() {
  for (int i = 1; i <= (N + 1); ++i) dp[i] = sp[i] = 0;
  return;
}
auto cmp = [](pair<int, int> A, pair<int, int> B) {
  if (A.second < B.second) return 1;
  return 0;
};
static inline int F(int byte) {
  if (!Restrictions[byte].size()) {
    int r = 1;
    for (int i = 1; i <= N; ++i)
      if (Mars[byte][i])
        continue;
      else
        r = (2LL * r) % (1LL * MOD);
    return r;
  }
  sort(Restrictions[byte].begin(), Restrictions[byte].end(), cmp);
  int pos = 0, Max_Left = 0, Max_Right = 0;
  for (int i = 1; i <= (N + 1); ++i) {
    while (pos < (int)Restrictions[byte].size() &&
           Restrictions[byte][pos].second == (i - 1))
      Max_Left = max(Max_Left, Restrictions[byte][pos].first),
      Max_Right = (i - 1), ++pos;
    sp[i] = sp[i - 1];
    if (Mars[byte][i] && i <= N) continue;
    dp[i] = sp[i - 1] - ((Max_Left > 1) ? sp[Max_Left - 1] : 0);
    dp[i] += ((dp[i] < 0) ? MOD : 0);
    if (!(Max_Right && Max_Right < i))
      dp[i] = (1LL * dp[i] + 1LL) % (1LL * MOD);
    sp[i] = (1LL * sp[i] + 1LL * dp[i]) % (1LL * MOD);
  }
  return dp[(N + 1)];
}
int main() {
  Read();
  Precalculation();
  long long ans = 1;
  for (int b = 0; b < K; ++b) {
    ans = (1LL * ans * F(b)) % (1LL * MOD);
    if (b != (K - 1)) Reset();
  }
  cout << ans << '\n';
  return 0;
}
