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;
void fastIO() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
template <typename T, typename U>
inline void mnze(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void mxze(T &x, U y) {
  if (x < y) x = y;
}
void _scan(int &x) { scanf("%d", &x); }
void _scan(long long &x) { scanf("%lld", &x); }
void _scan(double &x) { scanf("%lf", &x); }
void _scan(char &x) { scanf(" %c", &x); }
void _scan(char *x) { scanf("%s", x); }
template <typename T>
void _scan(vector<T> &v) {
  for (int i = 0; i < (int)v.size(); i++) _scan(v[i]);
}
void scan() {}
template <typename T, typename... U>
void scan(T &head, U &...tail) {
  _scan(head);
  scan(tail...);
}
template <typename T>
void _dbg(const char *sdbg, T h) {
  cerr << sdbg << "=" << h << "\n";
}
template <typename T, typename... U>
void _dbg(const char *sdbg, T h, U... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, t...);
}
long long MOD = 998244353;
inline long long Msum(long long x) { return x; }
template <typename... Rest>
inline long long Msum(long long x, Rest... rest) {
  return (x + Msum(rest...)) % MOD;
}
inline long long Mprod(long long x) { return x; }
template <typename... Rest>
inline long long Mprod(long long x, Rest... rest) {
  return x * Mprod(rest...) % MOD;
}
inline long long Mnorm(long long x) { return (x % MOD + MOD) % MOD; }
const int MAXN = 5e5 + 5;
int n, k, m;
vector<tuple<int, int, int>> restr;
int ones[MAXN], zeros[MAXN];
bool one[MAXN];
long long dp[MAXN];
long long dpp[MAXN];
int fR[MAXN];
long long solve(int bi) {
  for (int i = 0; i < n; i++) {
    ones[i] = -1;
    zeros[i] = n;
  }
  for (auto &[l, r, x] : restr) {
    if (x & (1 << bi))
      mxze(ones[l], r);
    else
      mnze(zeros[l], r);
  }
  int mxr1 = -1;
  for (int i = 0; i < n; i++) {
    mxze(mxr1, ones[i]);
    one[i] = (mxr1 >= i);
  }
  int curfR = n;
  fR[n] = curfR;
  for (int i = n - 1; i > -1; i--) {
    fR[i] = curfR;
    mnze(curfR, zeros[i]);
  }
  dpp[n] = dp[n] = 1;
  for (int i = n - 1; i > -1; i--) {
    if (one[i])
      dp[i] = 0;
    else
      dp[i] = ((dpp[i + 1] - dpp[fR[i]] + dp[fR[i]]) % MOD + MOD) % MOD;
    dpp[i] = (dp[i] + dpp[i + 1]) % MOD;
  }
  long long ans = 0;
  for (int i = 0; i < curfR + 1; i++) ans += dp[i];
  return ans % MOD;
}
int main() {
  scan(n, k, m);
  for (int i = 0; i < m; i++) {
    int l, r, x;
    scan(l, r, x);
    l--;
    r--;
    restr.push_back({l, r, x});
  }
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    ans = Mprod(ans, solve(i));
  }
  printf("%lld\n", ans);
}
