#include <bits/stdc++.h>
using namespace std;
const int N = (int)5e5 + 123, inf = 1e9, mod = 1e9 + 7, M = 511;
const long long INF = (long long)1e18;
int n, m, k, x, y, d, t, e;
long long A[M][M], a[2][M][M];
vector<pair<int, int> > ex[M][M], ey[M][M];
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(NULL);
  cin >> n >> m >> k;
  for (int i = 1; i <= k; ++i) {
    cin >> x >> y >> d >> t >> e;
    if (x + y >= t && ((x + y) - t) % 4 == 0) {
      A[x][y] += e;
      if (x + d + y >= t && ((x + d + y) - t) % 4 == 2) {
        ex[x + d][y].push_back(make_pair(d, e));
      }
      if (x + y + d >= t && ((x + y + d) - t) % 4 == 3) {
        ey[x][y + d].push_back(make_pair(d, e));
      }
    }
    if (x + d + y - d >= t && ((x + d + y - d) - t) % 4 == 1) {
      A[x + d][y - d] += e;
      if (x + d + y >= t && ((x + d + y) - t) % 4 == 2) {
        ey[x + d][y].push_back(make_pair(d, e));
      }
    }
    if (x + d + y >= t && ((x + d + y) - t) % 4 == 2) {
      A[x + d][y] += e;
    }
    if (x + y + d >= t && ((x + y + d) - t) % 4 == 3) {
      A[x][y + d] += e;
    }
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) a[0][i][j] = a[1][i][j] = INF;
  a[0][0][0] = a[1][0][0] = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      sort(ex[i][j].begin(), ex[i][j].end());
      sort(ey[i][j].begin(), ey[i][j].end());
    }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      long long cost = a[0][i][j];
      for (int k = 1; k + i < n; ++k) {
        cost += A[i + k][j];
        for (int w = 0; w < (int)ex[i + k][j].size(); ++w) {
          if (i + k - ex[i + k][j][w].first >= i)
            cost -= ex[i + k][j][w].second;
          else
            break;
        }
        a[1][i + k][j] = min(a[1][i + k][j], cost);
      }
      cost = a[1][i][j];
      for (int k = 1; k + j < m; ++k) {
        cost += A[i][j + k];
        for (int w = 0; w < (int)ey[i][j + k].size(); ++w) {
          if (j + k - ey[i][j + k][w].first >= j)
            cost -= ey[i][j + k][w].second;
          else
            break;
        }
        a[0][i][j + k] = min(a[0][i][j + k], cost);
      }
    }
  cout << min(a[0][n - 1][m - 1], a[1][n - 1][m - 1]);
  return 0;
}
