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 N = 5e5 + 5;
const int mod = 998244353;
int de[N], dp[N], f[N];
int sum[N];
struct node {
  int l, r;
  bool operator<(const node& a) const { return r < a.r; }
};
vector<node> s[30][2];
int cal(int, int);
void init();
void work();
int ca = 1;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  work();
  return 0;
}
void work() {
  int n, k, m;
  cin >> n >> k >> m;
  int l, r, x;
  while (m--) {
    cin >> l >> r >> x;
    for (int i = 0; i < k; i++) {
      if (x >> i & 1)
        s[i][1].push_back(node{l, r});
      else
        s[i][0].push_back(node{l, r});
    }
  }
  int ans = 1;
  for (int i = 0; i < k; i++) ans = 1LL * ans * cal(n, i) % mod;
  cout << ans;
}
int cal(int n, int k) {
  memset(de, 0, (n + 1) << 2);
  memset(dp, 0, (n + 1) << 2);
  for (auto t : s[k][1]) {
    de[t.l]++, de[t.r + 1]--;
  }
  for (int i = 1; i <= n; i++) de[i] += de[i - 1];
  int mx = 0;
  sort(s[k][0].begin(), s[k][0].end());
  for (int i = 1, j = 0; i <= n + 1; i++) {
    while (j < (int)s[k][0].size() && s[k][0][j].r < i) {
      mx = max(mx, s[k][0][j].l);
      j++;
    }
    f[i] = mx;
  }
  dp[0] = 1;
  sum[0] = 1;
  for (int i = 1; i <= n + 1; i++) {
    if (de[i] > 0)
      dp[i] = 0;
    else {
      if (f[i] < i) {
        if (f[i] == 0)
          dp[i] = sum[i - 1];
        else
          dp[i] = (sum[i - 1] - sum[f[i] - 1] + mod) % mod;
      }
    }
    sum[i] = (sum[i - 1] + dp[i]) % mod;
  }
  return dp[n + 1];
}
