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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx")
using namespace std;
template <typename T>
void uin(T &a, T b) {
  if (b < a) a = b;
}
template <typename T>
void uax(T &a, T b) {
  if (b > a) a = b;
}
const long long N = 500 * 1000 + 228;
const long long MOD = 998244353;
long long mod(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
struct seg {
  long long l, r, x;
  seg() {}
  seg(long long _l, long long _r, long long _x) { l = _l, r = _r, x = _x; }
};
long long n, k, m;
long long dp[N], maxl[N], pref[N], bal[N];
vector<seg> s;
long long calc(long long bit) {
  for (long long i = 0; i <= n + 2; ++i) {
    dp[i] = 0;
    maxl[i] = 0;
    pref[i] = 0;
    bal[i] = 0;
  }
  long long minR = n + 10;
  for (auto p : s) {
    if ((p.x >> bit) & 1) {
      ++bal[p.l];
      --bal[p.r + 1];
    } else {
      uax(maxl[p.r], p.l);
      uin(minR, p.r);
    }
  }
  for (long long i = 1; i <= n + 1; ++i) {
    uax(maxl[i], maxl[i - 1]);
  }
  for (long long i = 1; i <= n + 1; ++i) {
    bal[i] += bal[i - 1];
  }
  for (long long i = 1; i <= n + 1; ++i) {
    dp[i] = 0;
    if (bal[i])
      dp[i] = 0;
    else {
      dp[i] = mod(pref[i - 1] - pref[max(0LL, maxl[i - 1] - 1)]);
      if (minR >= i) ++dp[i];
    }
    pref[i] = mod(pref[i - 1] + dp[i]);
  }
  return dp[n + 1];
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k >> m;
  for (long long i = 0; i < m; ++i) {
    long long l, r, x;
    cin >> l >> r >> x;
    s.emplace_back(l, r, x);
  }
  long long answer = 1;
  for (long long bit = 0; bit < k; ++bit) {
    answer = mod(answer * calc(bit));
  }
  cout << answer << '\n';
  return 0;
}
