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 int mod = 998244353;
int add() { return 0; }
template <typename... M>
int add(int a, M... b) {
  int x = add(b...);
  return a + x - mod * (a + x >= mod);
}
int mul() { return 1; }
template <typename... M>
int mul(int a, M... b) {
  return int(1ll * a * mul(b...) % mod);
}
int pwr(int a, long long x) {
  return (x ? mul(pwr(mul(a, a), x >> 1), (x & 1 ? a : 1)) : 1);
}
int dvd(int a, int b) { return mul(a, pwr(b, mod - 2)); }
int sub(int a, int b) { return a - b + mod * (a < b); }
void inc(int& a, int b) { a = add(a, b); }
void dec(int& a, int b) { a = sub(a, b); }
const int K = 30, N = 5e5 + 10;
int mx[K][N], pr[K][N];
int dp[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n, k, m;
  cin >> n >> k >> m;
  memset(mx, -1, sizeof(mx));
  while (m--) {
    int l, r, x;
    cin >> l >> r >> x;
    --l, --r;
    for (int i = 0; i < k; ++i) {
      if ((x >> i) & 1) {
        ++pr[i][l];
        --pr[i][r + 1];
      } else {
        mx[i][r] = max(mx[i][r], l);
      }
    }
  }
  int ans = 1;
  for (int i = 0; i < k; ++i) {
    memset(dp, 0, sizeof(dp));
    dp[0] = 1;
    int sum = 1, ptr = 0;
    for (int j = 0; j < n; ++j) {
      pr[i][j + 1] += pr[i][j];
      if (pr[i][j] == 0) {
        dp[j + 1] = sum;
        sum = add(sum, sum);
      }
      while (ptr <= mx[i][j]) {
        dec(sum, dp[ptr++]);
      }
    }
    ans = mul(ans, sum);
  }
  cout << ans;
  return 0;
}
