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;
bitset<1048576> st[30];
vector<pair<int, int>> arr[500005];
int dp[500005], pre[500005];
int N, K, M;
void update(bitset<1048576>& s, int a, int b, int l, int r, int p) {
  if (s[p]) return;
  if (l > b || r < a) return;
  if (l >= a && r <= b) {
    s[p] = true;
    return;
  }
  int m = (l + r) / 2;
  update(s, a, b, l, m, 2 * p + 1);
  update(s, a, b, m + 1, r, 2 * p + 2);
}
void update(bitset<1048576>& s, int a, int b) { update(s, a, b, 0, 500005, 0); }
bool get(bitset<1048576>& s, int x, int l, int r, int p) {
  if (x >= l && x <= r && s[p]) return true;
  if (l == r && l == x) {
    return s[p];
  }
  int m = (l + r) / 2;
  if (x <= m) {
    return get(s, x, l, m, 2 * p + 1);
  }
  return get(s, x, m + 1, r, 2 * p + 2);
}
bool get(bitset<1048576>& s, int x) { return get(s, x, 0, 500005, 0); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> K >> M;
  for (int i = 0; i < M; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    arr[b].push_back({a, c});
    for (int j = 0; j < K; ++j) {
      bool cur = c & (1 << j);
      if (cur) {
        update(st[j], a, b);
      }
    }
  }
  long long ans = 1;
  for (int i = 0; i < K; ++i) {
    pre[0] = 0;
    if (!get(st[i], 0)) {
      dp[0] = 1;
      pre[1] = 1;
    } else {
      dp[0] = 0;
      pre[1] = 0;
    }
    int curS = -1;
    for (int j = 1; j < N; ++j) {
      for (int k = 0; k < arr[j - 1].size(); ++k) {
        if (arr[j - 1][k].second & (1 << i)) continue;
        curS = max(curS, arr[j - 1][k].first);
      }
      if (get(st[i], j)) {
        dp[j] = 0;
      } else {
        dp[j] = pre[j] - pre[max(0, curS)] + (curS == -1);
      }
      if (dp[j] < 0) dp[j] += 998244353;
      pre[j + 1] = pre[j] + dp[j];
      if (pre[j + 1] >= 998244353) pre[j + 1] -= 998244353;
    }
    for (int j = 0; j < arr[N - 1].size(); ++j) {
      if (arr[N - 1][j].second & (1 << i)) continue;
      curS = max(curS, arr[N - 1][j].first);
    }
    long long curA = pre[N] - pre[curS] + (curS == -1);
    if (curA < 0) curA += 998244353;
    ans = (1ll * ans * curA) % 998244353;
  }
  cout << ans << endl;
}
