#include<bits/stdc++.h>
#include<atcoder/all>
#define chmin(x,y) (x) = min((x),(y))
#define chmax(x,y) (x) = max((x),(y))
using namespace std;
using namespace atcoder;
using ll = long long;
const ll mod = 998244353;
using mint = modint998244353;
using Graph = vector<vector<int>>;
const vector<int> dx ={1,0,-1,0}, dy = {0,1,0,-1};

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;
}

int main() {
  // input
  int N; cin >> N;
  vector<ll> A(2*N),B(2*N);
  for(int i = 0; i < 2*N; i++) cin >> A[i];
  for(int i = 0; i < 2*N; i++) cin >> B[i];
  
  // solve: Kukan DP O(N^2)
  vector<vector<vector<ll>>> dp(N+1,vector<vector<ll>>(N+1,vector<ll>(3)));
  
  for(int i = 0; i < N; i++){
    if(A[2*i] - B[2*i+1] <= A[2*i+1] - B[2*i])
      dp[i][i+1] = {A[2*i] - B[2*i+1], 2*i, 2*i+1};
    else
      dp[i][i+1] = {A[2*i+1] - B[2*i], 2*i+1, 2*i};
  }
  
  for(int d = 2; d <= N; d++){
    for(int l = 0; l + d <= N; l++){
      int r = l+d;
      
      dp[l][r] = dp[l][r-1];
      dp[l][r][0] += dp[r-1][r][0];
      
      int swp1 = dp[l][r-1][2], swp2 = dp[r-1][r][1];
      ll X = dp[l][r][0] + B[swp1] - B[swp2] - A[swp2] + A[swp1];
      
      int swp3 = dp[r-1][r][2], swp4 = dp[l][r-1][1];
      ll Y = dp[l][r][0] + B[swp3] - B[swp4] - A[swp4] + A[swp3];
      if(min(X,Y) > dp[l][r][0]){
        if(min(X,Y) == X)
          dp[l][r] = {X,dp[l][r-1][1],swp2};
        else
          dp[l][r] = {Y,dp[r-1][r][1],swp4};
      }
      
    }
  }
  
  
  // output
  cout << dp[0][N][0] << endl;
}