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;
long long gi() {
  char cc = getchar();
  long long cn = 0, flus = 1;
  while (cc < '0' || cc > '9') {
    if (cc == '-') flus = -flus;
    cc = getchar();
  }
  while (cc >= '0' && cc <= '9') cn = cn * 10 + cc - '0', cc = getchar();
  return cn * flus;
}
const long long N = 5e5 + 5;
const long long P = 998244353;
long long n, m, k, a[N], f[N], Lx[N], ll[N], Nxt[N], Ans = 1;
struct S {
  long long l, r, x;
} q[N];
signed main() {
  n = gi(), k = gi(), m = gi();
  for (register long long i = 1; i <= m; ++i)
    q[i].l = gi(), q[i].r = gi(), q[i].x = gi();
  for (register long long l = 0; l < k; ++l) {
    long long L = -1, S = 0, ans = 0;
    for (register long long i = 1; i <= n; ++i) ll[i] = -1;
    for (register long long i = 1; i <= m; ++i) {
      if (q[i].x & (1ll << l))
        ++Lx[q[i].l], --Lx[q[i].r + 1];
      else
        ll[q[i].r] = max(ll[q[i].r], q[i].l - 1);
    }
    for (register long long i = 1; i <= n; ++i) S += Lx[i], a[i] = S ? 1 : 0;
    f[0] = 1, Nxt[0] = f[0];
    for (register long long i = 1; i <= n; ++i) {
      if (a[i])
        f[i] = 0;
      else
        f[i] = (Nxt[i - 1] - ((L >= 0) ? Nxt[L] : 0) + P) % P;
      L = max(L, ll[i]), Nxt[i] = (Nxt[i - 1] + f[i]) % P;
    }
    for (register long long i = L + 1; i <= n; ++i) ans = (ans + f[i]) % P;
    Ans = Ans * ans % P;
    for (register long long i = 1; i <= n + 1; ++i)
      f[i] = Nxt[i] = Lx[i] = a[i] = 0;
  }
  cout << Ans << endl;
  return 0;
}
