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 long long mod = 998244353;
template <class S, class T>
ostream& operator<<(ostream& os, const pair<S, T> v) {
  os << "(" << v.first << ", " << v.second << ")";
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << " ";
    }
    os << v[i];
  }
  return os;
}
template <class T>
ostream& operator<<(ostream& os, const vector<vector<T>> v) {
  for (int i = 0; i < (int)v.size(); i++) {
    if (i > 0) {
      os << endl;
    }
    os << v[i];
  }
  return os;
}
template <class T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
string num2bit(long long num, long long len) {
  string bit = "";
  for (int i = 0; i < (int)len; ++i) {
    bit += char('0' + (num >> i & 1));
  }
  return bit;
}
struct seg {
  long long l, r, x;
};
vector<seg> S;
long long n, k, m;
long long calc(int pos) {
  vector<seg> T;
  vector<long long> imos(n + 1, 0);
  for (auto se : S) {
    if (se.x >> pos & 1) {
      imos[se.l]++;
      imos[se.r + 1]--;
    } else {
      T.push_back(se);
    }
  }
  for (int i = 0; i < (int)n; ++i) imos[i + 1] += imos[i];
  sort(T.begin(), T.end(), [](seg a, seg b) { return a.r < b.r; });
  vector<long long> dp(n + 1, 0);
  vector<long long> Sdp(n + 2, 0);
  dp[0] = 1;
  long long L = -1;
  int idx = 0;
  Sdp[1] = dp[0];
  for (int i = 1; i <= n; i++) {
    while (1) {
      if (idx >= (int)T.size() or T[idx].r > i - 2) break;
      chmax(L, T[idx].l);
      idx++;
    }
    if (imos[i - 1] == 0) {
      dp[i] = Sdp[i] - Sdp[L + 1];
      dp[i] += mod;
      dp[i] %= mod;
    }
    Sdp[i + 1] = Sdp[i] + dp[i];
    Sdp[i + 1] %= mod;
  }
  while (idx < (int)T.size()) {
    chmax(L, T[idx].l);
    idx++;
  }
  long long res = 0;
  for (int i = L + 1; i <= n; i++) {
    res += dp[i];
    res %= mod;
  }
  return res;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> k >> m;
  for (int i = 0; i < (int)m; ++i) {
    long long l, r, x;
    cin >> l >> r >> x;
    l--, r--;
    S.push_back((seg){l, r, x});
  }
  long long res = 1;
  for (int i = 0; i < (int)k; ++i) {
    res *= calc(i);
    res %= mod;
  }
  cout << res << endl;
  return 0;
}
