#include using namespace std; #include using namespace atcoder; template inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); } template inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); } #define rep(i, n) for (long long i = 0; i < (long long)(n); i++) #define rep2(i, m ,n) for (int i = (m); i < (long long)(n); i++) #define REP(i, n) for (long long i = 1; i < (long long)(n); i++) typedef long long ll; #define updiv(N,X) (N + X - 1) / X #define l(n) n.begin(),n.end() #define YesNo(Q) Q==1?cout<<"Yes":cout<<"No" using P = pair; using mint = modint; const int MOD = 998244353LL; const ll INF = 999999999999LL; vector fact, fact_inv, inv; /* init_nCk :二項係数のための前処理 計算量:O(n) */ template void input(vector &v){ rep(i,v.size()){cin>>v[i];} return; } void init_nCk(int SIZE) { fact.resize(SIZE + 5); fact_inv.resize(SIZE + 5); inv.resize(SIZE + 5); fact[0] = fact[1] = 1; fact_inv[0] = fact_inv[1] = 1; inv[1] = 1; for (int i = 2; i < SIZE + 5; i++) { fact[i] = fact[i - 1] * i % MOD; inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD; fact_inv[i] = fact_inv[i - 1] * inv[i] % MOD; } } /* nCk :MODでの二項係数を求める(前処理 int_nCk が必要) 計算量:O(1) */ long long nCk(int n, int k) { assert(!(n < k)); assert(!(n < 0 || k < 0)); return fact[n] * (fact_inv[k] * fact_inv[n - k] % MOD) % MOD; } long long modpow(long long a, long long n, long long mod) { long long res = 1; while (n > 0) { if (n & 1) res = res * a % mod; a = a * a % mod; n >>= 1; } return res; } ll POW(ll a,ll n){ long long res = 1; while (n > 0) { if (n & 1) res = res * a; a = a * a; n >>= 1; } return res; } int main() { int n;cin>>n;n *= 2; string s;cin>>s; ll dp[n+1][n+1]; rep(i,n+1){rep(j,n+1){dp[i][j]=999999999999LL;}} vector v(n); rep(i,n){cin>>v[i];} dp[0][0] = 0; rep(i,n){ rep(j,n){ if(j==0){ if(s[i]=='('){dp[i+1][j+1] = min(dp[i][j],dp[i+1][j+1]);} else{dp[i+1][j+1] = min(dp[i][j]+v[i],dp[i+1][j+1]);} } else{ if(s[i]=='('){dp[i+1][j+1] = min(dp[i][j],dp[i+1][j+1]); dp[i+1][j-1] = min(dp[i][j]+v[i],dp[i+1][j-1]); } else{ dp[i+1][j+1] = min(dp[i][j]+v[i],dp[i+1][j+1]); dp[i+1][j-1] = min(dp[i][j],dp[i+1][j-1]); } } }} cout << dp[n][0] << endl; }