Problem: Consider a square grid with h rows and w columns with some dominoes on it. Each domino covers exactly two cells of the grid that share a common side. Every cell is covered by at most one domino.

Let's call a placement of dominoes on the grid perfectly balanced if no row and no column contains a pair of cells covered by two different dominoes. In other words, every row and column may contain no covered cells, one covered cell, or two covered cells that belong to the same domino.

You are given a perfectly balanced placement of dominoes on a grid. Find the number of ways to place zero or more extra dominoes on this grid to keep the placement perfectly balanced. Output this number modulo 998 244 353.

Input

The first line contains three integers h, w, and n (1 ≤ h, w ≤ 3600; 0 ≤ n ≤ 2400), denoting the dimensions of the grid and the number of already placed dominoes. The rows are numbered from 1 to h, and the columns are numbered from 1 to w.

Each of the next n lines contains four integers r_{i, 1}, c_{i, 1}, r_{i, 2}, c_{i, 2} (1 ≤ r_{i, 1} ≤ r_{i, 2} ≤ h; 1 ≤ c_{i, 1} ≤ c_{i, 2} ≤ w), denoting the row id and the column id of the cells covered by the i-th domino. Cells (r_{i, 1}, c_{i, 1}) and (r_{i, 2}, c_{i, 2}) are distinct and share a common side.

The given domino placement is perfectly balanced.

Output

Output the number of ways to place zero or more extra dominoes on the grid to keep the placement perfectly balanced, modulo 998 244 353.

Examples

Input


5 7 2
3 1 3 2
4 4 4 5


Output


8


Input


5 4 2
1 2 2 2
4 3 4 4


Output


1


Input


23 42 0


Output


102848351

Note

In the first example, the initial grid looks like this:

<image>

Here are 8 ways to place zero or more extra dominoes to keep the placement perfectly balanced:

<image>

In the second example, the initial grid looks like this:

<image>

No extra dominoes can be placed here.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(static_cast<unsigned int>(
    chrono::steady_clock::now().time_since_epoch().count()));
template <typename num_t>
inline void add_mod(num_t& a, const long long& b, const int& m) {
  a = (a + b) % m;
  if (a >= m) a -= m;
  if (a < 0) a += m;
}
template <typename num_t>
inline bool update_max(num_t& a, const num_t& b) {
  return a < b ? a = b, true : false;
}
template <typename num_t>
inline bool update_min(num_t& a, const num_t& b) {
  return a > b ? a = b, true : false;
}
template <typename num_t>
num_t gcd(num_t lhs, num_t rhs) {
  return !lhs ? rhs : gcd(rhs % lhs, lhs);
}
template <typename num_t>
num_t pw(num_t n, num_t k, const num_t& mod) {
  if (k == -1) k = mod - 2;
  num_t res = 1;
  for (; k > 0; k >>= 1) {
    if (k & 1) res = 1ll * res * n % mod;
    n = 1ll * n * n % mod;
  }
  return res % mod;
}
const int inf = 1e9 + 7;
const int mod = 998244353;
const long long ll_inf = (long long)inf * inf + 7;
const int maxn = 4000 + 7;
struct InverseModule {
  static const int MOD = ::mod;
  static const int MAX_N = 1e6 + 7;
  int fact[MAX_N];
  int inv_fact[MAX_N];
  InverseModule() {
    fact[0] = 1;
    for (int i = 1; i < MAX_N; ++i) {
      fact[i] = (long long)fact[i - 1] * i % MOD;
    }
    inv_fact[MAX_N - 1] = inverse(fact[MAX_N - 1]);
    for (int i = MAX_N - 2; i >= 0; --i) {
      inv_fact[i] = (long long)inv_fact[i + 1] * (i + 1) % MOD;
    }
  }
  int inverse(int x, int mod = MOD) {
    int r = mod, newr = x;
    int t = 0, newt = 1;
    while (newr > 0) {
      int q = r / newr;
      int tmp = newr;
      newr = r % newr;
      r = tmp;
      tmp = newt;
      newt = t - q * newt;
      t = tmp;
    }
    if (t < 0) t += mod;
    return t;
  }
  int choose(int n, int k) {
    return (long long)fact[n] * inv_fact[n - k] % MOD * inv_fact[k] % MOD;
  }
  int perm(int n, int k) { return (long long)fact[n] * inv_fact[n - k] % MOD; }
  long long pw(long long x, long long k, const long long& mod = MOD) {
    long long res = 1 % mod;
    for (; k > 0; k >>= 1) {
      if (k & 1) res = res * x % mod;
      x = x * x % mod;
    }
    return res;
  }
  pair<int, int> extended_euclid(int a, int b) {
    int s0 = 1, t0 = 0;
    int s1 = 0, t1 = 1;
    while (b > 0) {
      int q = a / b;
      int tmp = 0;
      tmp = s1;
      s1 = s0 - q * s1;
      s0 = tmp;
      tmp = t1;
      t1 = t0 - q * t1;
      t0 = tmp;
      tmp = b;
      b = a - q * b;
      a = tmp;
    }
    return {s0, t0};
  }
  int stirling_second_kind(int n, int k) {
    long long res = 0;
    for (int j = 0; j <= k; ++j) {
      long long temp = 1ll * choose(k, j) * pw(j, n);
      if ((k - j) & 1)
        res = (res - temp + MOD) % MOD;
      else
        res = (res + temp) % MOD;
    }
    return res * inv_fact[k] % MOD;
  }
} inv;
int n, m, k;
long long dp[2][maxn];
long long f[maxn][maxn];
void pre_calc() {
  f[0][0] = 1;
  for (int len = 1; len < maxn; ++len) {
    for (int num_brick = (0), _b = (maxn); num_brick < _b; ++num_brick) {
      f[len][num_brick] = f[len - 1][num_brick];
      if (len >= 2 && num_brick > 0)
        add_mod(f[len][num_brick], f[len - 2][num_brick - 1], mod);
    }
  }
}
vector<long long> calc(vector<bool> arr) {
  int cur = 0;
  for (int i = (0), _b = (maxn); i < _b; ++i) dp[cur][i] = 0;
  dp[cur][0] = 1;
  for (int i = 0; i < arr.size(); ++i)
    if (arr[i]) {
      int nxt_i = i;
      while (nxt_i < arr.size() && arr[nxt_i]) ++nxt_i;
      int len = nxt_i - i;
      i = nxt_i - 1;
      if (len < 2) continue;
      int pre = cur;
      cur ^= 1;
      for (int i = (0), _b = (maxn); i < _b; ++i) dp[cur][i] = dp[pre][i];
      for (int d = 1; d * 2 <= len; ++d) {
        for (int pre_d = 0; pre_d < maxn; ++pre_d)
          if (dp[pre][pre_d]) {
            add_mod(dp[cur][d + pre_d], dp[pre][pre_d] * f[len][d], mod);
          }
      }
    }
  return vector<long long>(dp[cur], dp[cur] + maxn);
}
void solve() {
  pre_calc();
  cin >> n >> m >> k;
  vector<bool> valid_row(n, true), valid_col(m, true);
  for (int i = (0), _b = (k); i < _b; ++i) {
    int u1, v1, u2, v2;
    cin >> u1 >> v1 >> u2 >> v2;
    valid_row[--u1] = false;
    valid_row[--u2] = false;
    valid_col[--v1] = false;
    valid_col[--v2] = false;
  }
  int total_row = 0;
  int total_col = 0;
  for (int i = (0), _b = (n); i < _b; ++i) total_row += valid_row[i];
  for (int j = (0), _b = (m); j < _b; ++j) total_col += valid_col[j];
  vector<long long> lhs = calc(valid_row), rhs = calc(valid_col);
  long long res = 0;
  for (int a = 0; a < maxn; ++a)
    if (lhs[a])
      for (int b = 0; b < maxn; ++b)
        if (rhs[b]) {
          if (2 * a + b > total_row || a + 2 * b > total_col) break;
          long long cur = lhs[a] * rhs[b] % mod;
          cur = cur * inv.choose(total_row - a * 2, b) % mod;
          cur = cur * inv.choose(total_col - b * 2, a) % mod;
          cur = cur * inv.fact[a] % mod * inv.fact[b] % mod;
          add_mod(res, cur, mod);
        }
  cout << res << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  const bool multiple_test = false;
  int test = 1;
  if (multiple_test) cin >> test;
  for (int i = 0; i < test; ++i) {
    solve();
  }
}