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;
int n, k, m;
const int maxn = 5e5 + 100;
const int mod = 998244353;
struct node {
  long long l, r, x;
} save[maxn];
long long dp[maxn], sum[maxn], mx[maxn], sub[maxn];
int main() {
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%lld%lld%lld", &save[i].l, &save[i].r, &save[i].x);
  }
  long long res = 1;
  for (int kk = 0; kk < k; kk++) {
    memset(sub, 0, sizeof(sub));
    memset(mx, 0, sizeof(mx));
    memset(dp, 0, sizeof(dp));
    memset(sum, 0, sizeof(sum));
    dp[0] = 1;
    sum[0] = 1;
    int tmp = 1 << kk;
    for (int i = 1; i <= m; i++) {
      if (save[i].x & tmp) {
        sub[save[i].l]++;
        sub[save[i].r + 1]--;
      } else {
        mx[save[i].r] = max(mx[save[i].r], save[i].l);
      }
    }
    for (int i = 1; i <= n; i++) sub[i] = sub[i] + sub[i - 1];
    int p = 0;
    for (int i = 1; i <= n; i++) {
      sum[i] = sum[i - 1];
      if (sub[i] == 0) {
        dp[i] = sum[i - 1] % mod;
        sum[i] += dp[i];
        sum[i] %= mod;
      }
      while (p < mx[i]) sum[i] = (sum[i] - dp[p]) % mod, p++;
    }
    res *= sum[n];
    res %= mod;
    res += mod;
    res %= mod;
  }
  printf("%I64d", res);
  return 0;
}
