#include <iostream>
using namespace std;
const int N = 505;
int a[N][N],dp[N][N];
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i ++ ){
        for (int j = 0; j <= n+1; j ++ ){
            a[i][j]= -0x3f3f3f3f3f3f;
        }
    }

    for (int i = 1; i <= n; i ++ ){
        for (int j = 1; j <= i; j++ ){
            cin >> a[i][j];
        }
    }
    /*
    for (int i = n; i >=1; i -- ){ //从底 n 行往上推
        for (int j = 1; j <= i; j ++ ){
            a[i][j]+=max(a[i+1][j+1],a[i+1][j]);
        }
    }
    cout << a[1][1] << endl;
    */
    // 从上往下走
    for(int i=1;i<=n;i++)
        for(int j=1;j<=i;j++) {
            cin>>a[i][j];
            a[i][j]+=max(a[i-1][j],a[i-1][j-1]);
        }
    int ans = -0x3f3f3f3f3f3f;

    for (int i = 1; i <= n; i ++ ){
        //cout << a[n][i] << endl;
        ans = max(ans,a[n][i]); //由于从上往下,最后一层要判断走哪个元素最大
    }

    cout << ans << endl;
    // dp[i][j] 定义为第i行，走到第j列的最大路径和。
    // dp[i][j] 可以从左边下来，也可以从右边下来。dp[i][j]=max(dp[i-1][j-1]+a[i-1][j-1],dp[i-1][j]+a[i-1][j]);
    return 0;
}