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 double eps = 1e-9;
const int inf = 2000000000;
const long long infLL = 9000000000000000000;
template <typename first, typename second>
ostream& operator<<(ostream& os, const pair<first, second>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ",";
    os << *it;
  }
  return os << "]";
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream& operator<<(ostream& os, const map<first, second>& v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
void faltu() { cerr << '\n'; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << '\n';
}
template <typename T, typename... hello>
void faltu(T arg, const hello&... rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
inline bool checkBit(long long n, int i) { return n & (1LL << i); }
inline long long setBit(long long n, int i) {
  return n | (1LL << i);
  ;
}
inline long long resetBit(long long n, int i) { return n & (~(1LL << i)); }
const int mx = 5e5 + 5;
int n, k, m;
int l[mx], r[mx], x[mx];
long long dp[mx];
long long pre[mx];
int one[mx];
int f[mx];
long long calc(int pos) {
  memset(dp, 0, sizeof(dp));
  memset(one, 0, sizeof(one));
  memset(f, 0, sizeof(f));
  for (int i = 0; i < m; ++i) {
    if (checkBit(x[i], pos)) {
      ++one[l[i]];
      --one[r[i] + 1];
    } else {
      f[r[i]] = max(f[r[i]], l[i]);
    }
  }
  for (int i = 1; i < mx; ++i) {
    one[i] += one[i - 1];
    f[i] = max(f[i], f[i - 1]);
  }
  bool ok = 0;
  dp[0] = 1;
  pre[0] = 1;
  for (int i = 1; i <= n + 1; ++i) {
    if (one[i]) {
      pre[i] = pre[i - 1];
      continue;
    }
    if (f[i - 1] - 1 >= 0)
      dp[i] = pre[i - 1] - pre[f[i - 1] - 1];
    else
      dp[i] = pre[i - 1];
    pre[i] = dp[i] + pre[i - 1];
    pre[i] %= 998244353;
    dp[i] %= 998244353;
  }
  return dp[n + 1];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k >> m;
  for (int i = 0; i < m; ++i) {
    cin >> l[i] >> r[i] >> x[i];
  }
  long long ans = 1;
  for (int i = 0; i < k; ++i) {
    ans *= calc(i);
    ans %= 998244353;
    ans += 998244353;
    ans %= 998244353;
  }
  cout << ans << '\n';
}
