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")
using namespace std;
const long long mod0 = 1e9 + 7;
const long long mod1 = 998244353;
const long long mod2 = 1e9 + 9;
const long long mod3 = 2147483647;
const long long inf0 = 2e9;
const int inf = 1e9;
const long long INF = 2e18;
const long double eps = 1e-15;
void init() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int maxn = 5e5 + 5;
int n, k, m;
vector<vector<int> > Q;
long long calc(int t) {
  vector<long long> ones(maxn), dp(maxn);
  vector<int> mx(maxn, -1);
  for (int i = 0; i < m; ++i) {
    int l = Q[i][0], r = Q[i][1];
    if (Q[i][2] & (1 << t)) {
      ones[l]++;
      ones[r + 1]--;
    } else {
      mx[r] = max(mx[r], l);
    }
  }
  int j = -1;
  for (int i = 0; i < n; ++i) {
    long long cur = 0;
    if (!ones[i]) {
      cur = dp[i];
      if (j == -1)
        cur++;
      else
        cur -= dp[j];
    }
    dp[i + 1] = (dp[i] + cur) % mod1;
    if (dp[i + 1] < 0) dp[i + 1] += mod1;
    ones[i + 1] += ones[i];
    j = max(j, mx[i]);
  }
  long long ans = dp[n];
  if (j == -1)
    ans++;
  else
    ans -= dp[j];
  ans %= mod1;
  if (ans < 0) ans += mod1;
  return ans;
}
void solve() {
  cin >> n >> k >> m;
  Q.resize(m, vector<int>(3));
  for (int i = 0; i < m; ++i) {
    cin >> Q[i][0] >> Q[i][1] >> Q[i][2];
    Q[i][0]--, Q[i][1]--;
  }
  long long ans = 1;
  for (int i = 0; i < k; ++i) ans = (ans * calc(i)) % mod1;
  cout << ans << '\n';
}
bool multitest = false;
int main() {
  init();
  int t = 1;
  if (multitest) cin >> t;
  for (int i = 0; i < t; ++i) solve();
}
