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;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
const long long MOD = 998244353;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
long long ans = 1;
vector<pair<pair<long long, long long>, long long> > qr;
long long n, m, k;
long long ex[500005], rng[500005], dp[500005], dp2[500005], dp3[500005],
    sdp[500005], pf[500005];
void solve(long long bit) {
  vector<pair<pair<long long, long long>, long long> > Q;
  for (long long i = 1; i <= n; i++)
    ex[i] = rng[i] = dp[i] = dp2[i] = dp3[i] = sdp[i] = pf[i] = 0;
  for (auto q : qr) {
    Q.push_back({q.first, !(q.second & (1 << bit - 1))});
  }
  for (auto q : Q) {
    if (!q.second) {
      pf[q.first.first]++;
      pf[q.first.second + 1]--;
    }
  }
  long long mxid = 0;
  dp2[0] = 0, dp3[n + 1] = n + 1;
  for (long long i = 1; i <= n; i++) {
    pf[i] += pf[i - 1];
    if (!pf[i]) ex[i] = ++mxid;
    dp2[i] = !pf[i] ? i : dp2[i - 1];
  }
  for (long long i = n; i >= 1; i--) dp3[i] = !pf[i] ? i : dp3[i + 1];
  for (auto q : Q) {
    if (q.second) {
      long long l = dp3[q.first.first], r = dp2[q.first.second];
      if (l > r) {
        ans = 0;
        return;
      }
      rng[ex[r]] = ex[l];
    }
  }
  for (long long i = 1; i <= mxid; i++) rng[i] = max(rng[i], rng[i - 1]);
  dp[0] = sdp[0] = 1;
  for (long long i = 1; i <= mxid; i++) {
    dp[i] = (sdp[i - 1] - (rng[i - 1] == 0 ? 0 : sdp[rng[i - 1] - 1])) % MOD;
    sdp[i] = (sdp[i - 1] + dp[i]) % MOD;
  }
  long long re =
      mxid == 0 ? 1
                : (sdp[mxid] - (rng[mxid] == 0 ? 0 : sdp[rng[mxid] - 1])) % MOD;
  ans = ans * re % MOD;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k >> m;
  for (long long i = 1; i <= m; i++) {
    long long l, r, val;
    cin >> l >> r >> val;
    qr.push_back({{l, r}, val});
  }
  sort(begin(qr), end(qr));
  for (long long i = 1; i <= k; i++) solve(i);
  while (ans < 0) ans += MOD;
  cout << ans;
}
