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 MAXN = 500010;
const int mo = 998244353;
struct node {
  int l, r, x;
  inline bool operator<(const node &p) const { return r < p.r; }
} a[MAXN];
int n, k, m;
int isright[MAXN], last[MAXN], maxle[MAXN];
int one[MAXN], dp[MAXN], dpa[MAXN];
inline int read() {
  int x = 0, f = 1;
  char ch = 0;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  return x * f;
}
int main() {
  n = read();
  k = read();
  m = read();
  for (int i = 1; i <= m; ++i) {
    a[i].l = read(), a[i].r = read(), a[i].x = read();
  }
  if (n == 1 && m > 1) {
    for (int i = 1; i <= m - 1; ++i)
      if (a[i].x != a[i + 1].x) return puts("0"), 0;
  }
  sort(a + 1, a + m + 1);
  int ans = 1;
  for (int bit = 0; bit < k; ++bit) {
    memset(isright, 0, sizeof(isright));
    memset(one, 0, sizeof(one));
    memset(last, 0, sizeof(last));
    memset(dp, 0, sizeof(0));
    memset(dpa, 0, sizeof(dpa));
    for (int i = 1; i <= m; ++i)
      if ((a[i].x & (1 << bit))) one[a[i].l]++, one[a[i].r + 1]--;
    int maxL = 0;
    for (int h = 1; h <= m;)
      for (int i = 1; i <= n; ++i) {
        isright[i] = maxL;
        while (a[h].r == i) {
          if (!(a[h].x & (1 << bit))) maxL = max(maxL, a[h].l);
          h++;
        }
      }
    for (int i = 1; i <= n; ++i) one[i] += one[i - 1];
    if (one[n] == n) continue;
    dp[0] = dpa[0] = 1;
    for (int i = 1; i <= n; ++i) {
      if (!one[i]) {
        if (isright[i]) {
          dp[i] = (dpa[i - 1] - (dpa[isright[i] - 1]) + mo) % mo;
          dpa[i] = (dp[i] + dpa[i - 1]) % mo;
        } else
          dp[i] = dpa[i - 1], dpa[i] = (dp[i] + dpa[i - 1]) % mo;
      } else
        dp[i] = 0, dpa[i] = dpa[i - 1];
    }
    ans = 1LL * ans * (dpa[n] - dpa[maxL - 1] + mo) % mo;
  }
  printf("%d\n", ans);
}
