#include <bits/stdc++.h> using namespace std; // O(N^3) DP for JOI Heat (for teaching purposes, NOT for OJ) const int INF = 1e9; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int L, N; cin >> L; vector<int> C(L+2); for (int i = 1; i <= L; i++) cin >> C[i]; cin >> N; vector<int> X(N+1); for (int i = 1; i <= N; i++) cin >> X[i]; // For each road i: store the times of patients vector<vector<int>> tim(L+2); for (int t = 1; t <= N; t++) { int r = X[t]; tim[r].push_back(t); } // dp[i][j][t] : after processing road i, // sent j people to the right on road i, // and BV i+1 becomes full at time t (t = N+1 => never full) // dp value = max helicopters vector<vector<vector<int>>> dp(L+2, vector<vector<int>>(N+2, vector<int>(N+3, -INF))); // Base case: before processing any road dp[0][0][N+1] = 0; // Process each road i from 1..L-1 for (int i = 1; i <= L-1; i++) { int Z = tim[i].size(); // number of patients on this road // Precompute prefix count: // cnt[k] = number of patients with time <= tim[i][k-1] vector<int> cnt(Z+1); for (int k = 1; k <= Z; k++) cnt[k] = k; // Try all j = number sent right on this road for (int j = 0; j <= Z; j++) { // You send j to the right, Z-j to the left for (int t = 1; t <= N+1; t++) { int best = dp[i-1][j][t]; if (best < -1e8) continue; // left hospital = i, right hospital = i+1 // It has capacity C[i] and C[i+1] // Let’s decide how many left we send to each hospital: // left_count = Z - j // right_count = j // Compute time BV i becomes full: int left_need = C[i]; int right_need = C[i+1]; int T_left = N+1, T_right = N+1; if (Z - j >= C[i]) T_left = tim[i][C[i]-1]; if (j >= C[i+1]) T_right = tim[i][j-1]; int full_time = max(T_left, T_right); // helicopters = number of patients whose time > full_time int heli = 0; for (int k = 0; k < Z; k++) if (tim[i][k] > full_time) heli++; // Now update dp[i][j][t_new], // where t_new = T_right (time BV i+1 becomes full) int t_new = T_right; dp[i][j][t_new] = max(dp[i][j][t_new], best + heli); } } } // Final answer = max dp[L-1][0][t] int ans = 0; for (int t = 1; t <= N+1; t++) ans = max(ans, dp[L-1][0][t]); cout << ans << "\n"; return 0; }