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;
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream& operator<<(ostream& os, const T_container& v) {
  string sep;
  for (const T& x : v) os << sep << x, sep = " ";
  return os;
}
const int mxN = 5e5 + 10, oo = 1e9;
const long long MOD = 998244353;
class mint {
 public:
  int d;
  mint() { d = 0; }
  mint(long long _d) : d(_d % MOD){};
  mint operator*(const mint& o) const { return ((long long)d * o.d) % MOD; }
  mint operator+(const mint& o) const {
    long long tmp = d + o.d;
    if (tmp >= MOD) tmp -= MOD;
    return tmp;
  }
  mint operator-(const mint& o) const {
    long long tmp = d - o.d;
    if (tmp < 0) tmp += MOD;
    return tmp;
  }
  mint operator^(long long b) const {
    mint tmp = 1;
    mint power = *this;
    while (b) {
      if (b & 1) {
        tmp = tmp * power;
      }
      power = power * power;
      b /= 2;
    }
    return tmp;
  }
  mint operator/(const mint& o) { return *this * (o ^ (MOD - 2)); }
  bool operator==(const mint& o) { return d == o.d; }
  friend mint& operator+=(mint& a, const mint& o) {
    a.d += o.d;
    if (a.d >= MOD) a.d -= MOD;
    return a;
  }
};
int one[mxN], mx[mxN];
mint dp[mxN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k, m;
  cin >> n >> k >> m;
  vector<array<int, 3> > hs(m);
  for (auto& [l, r, x] : hs) cin >> l >> r >> x;
  mint res = 1;
  vector<mint> pw(n + 10);
  pw[0] = 1;
  for (int i = 1; i < pw.size(); ++i) {
    pw[i] = pw[i - 1] * 2;
  }
  for (int b = 0; b < k; ++b) {
    fill(one, one + n + 2, 0);
    fill(mx, mx + n + 2, 0);
    for (const auto [l, r, x] : hs) {
      if (1 << b & x) {
        one[l]++;
        one[r + 1]--;
      }
    }
    for (int i = 1; i <= n + 1; ++i) {
      one[i] += one[i - 1];
      one[i - 1] = !one[i - 1];
    }
    for (const auto [l, r, x] : hs) {
      if (!(1 << b & x)) {
        mx[r] = max(mx[r], l);
      }
    }
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
      mx[i] = max(mx[i], mx[i - 1]);
      dp[i] = 0;
      if (one[i]) {
        dp[i] = dp[i - 1];
        if (mx[i - 1]) dp[i] = dp[i] - dp[mx[i - 1] - 1];
      }
      dp[i] += dp[i - 1];
    }
    auto fact = dp[n];
    if (mx[n]) fact = fact - dp[mx[n] - 1];
    res = res * fact;
  }
  cout << res.d << '\n';
}
