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("O2")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx,avx2,sse,sse2,ssse3,tune=native")
using namespace std;
using ll = long long;
using pi = pair<int, int>;
using vi = vector<int>;
using vpi = vector<pi>;
using ld = long double;
const int maxn = 1 << 19, mod = (119 << 23) + 1, lg = 19;
int n, k, m, dp[maxn], pref[maxn], forced[maxn], fend[maxn];
vector<array<int, 3>> q;
int solve(int K) {
  memset(fend, 0, sizeof fend);
  memset(forced, 0, sizeof forced);
  memset(dp, 0, sizeof dp);
  memset(pref, 0, sizeof pref);
  for (auto i : q) {
    if ((i[2] >> K) & 1) {
      forced[i[0]]++;
      forced[i[1] + 1]--;
    } else {
      fend[i[1] + 1] = max(fend[i[1] + 1], i[0]);
    }
  }
  for (int i = 1; i <= n + 1; i++) {
    fend[i] = max(fend[i], fend[i - 1]);
    forced[i] += forced[i - 1];
  }
  dp[0] = pref[0] = 1;
  for (int i = 1; i <= n; i++) {
    pref[i] = pref[i - 1];
    if (forced[i]) continue;
    if (fend[i] >= 1)
      dp[i] = (mod + pref[i] - pref[fend[i] - 1]) % mod;
    else
      dp[i] = pref[i];
    pref[i] += dp[i];
    if (pref[i] >= mod) pref[i] -= mod;
  }
  int ans = 0;
  for (int i = n; i >= fend[n + 1]; i--) {
    ans += dp[i];
    if (ans >= mod) ans -= mod;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k >> m;
  q.resize(m);
  for (auto &i : q) cin >> i[0] >> i[1] >> i[2];
  ll ans = 1;
  for (int i = 0; i < k; i++) ans = ans * 1ll * solve(i) % mod;
  cout << ans;
  return 0;
}
