You are given a rectangular table 3 × n. Each cell contains an integer. You can move from one cell to another if they share a side.

Find such path from the upper left cell to the bottom right cell of the table that doesn't visit any of the cells twice, and the sum of numbers written in the cells of this path is maximum possible.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of columns in the table.

Next three lines contain n integers each — the description of the table. The j-th number in the i-th line corresponds to the cell aij ( - 109 ≤ aij ≤ 109) of the table.

Output

Output the maximum sum of numbers on a path from the upper left cell to the bottom right cell of the table, that doesn't visit any of the cells twice.

Examples

Input

3
1 1 1
1 -1 1
1 1 1


Output

7


Input

5
10 10 10 -1 -1
-1 10 10 10 10
-1 10 10 10 10


Output

110

Note

The path for the first example:

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The path for the second example:

<image>
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
inline long long add(long long _x, long long _y,
                     long long _mod = 1000000007LL) {
  _x += _y;
  return _x >= _mod ? _x - _mod : _x;
}
inline long long sub(long long _x, long long _y,
                     long long _mod = 1000000007LL) {
  _x -= _y;
  return _x < 0 ? _x + _mod : _x;
}
inline long long mul(long long _x, long long _y,
                     long long _mod = 1000000007LL) {
  _x *= _y;
  return _x >= _mod ? _x % _mod : _x;
}
long long mypow(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 1LL;
  long long _ret = mypow(mul(_a, _a, _mod), _x >> 1, _mod);
  if (_x & 1) _ret = mul(_ret, _a, _mod);
  return _ret;
}
long long mymul(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 0LL;
  long long _ret = mymul(add(_a, _a, _mod), _x >> 1, _mod);
  if (_x & 1) _ret = add(_ret, _a, _mod);
  return _ret;
}
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
int __ = 1, _cs;
void build() {}
long long n, a[3][101010], dp[101010][4];
void init() {
  n = getint();
  for (int j = 0; j < 3; j++)
    for (int i = 1; i <= n; i++) a[j][i] = getint();
}
inline long long sum(int col, int fr, int to) {
  if (fr > to) swap(fr, to);
  long long ret = 0;
  for (int i = fr; i <= to; i++) ret += a[i][col];
  return ret;
}
void solve() {
  dp[0][0] = 0;
  dp[0][1] = -10000000000000000LL;
  dp[0][2] = -10000000000000000LL;
  dp[0][3] = -10000000000000000LL;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 4; j++) dp[i][j] = -10000000000000000LL;
    for (int j = 0; j < 3; j++)
      for (int nj = 0; nj < 3; nj++)
        dp[i][nj] = max(dp[i][nj], dp[i - 1][j] + sum(i, j, nj));
    long long al = sum(i, 0, 2);
    for (int j : {0, 2}) {
      dp[i][3] = max(dp[i][3], dp[i - 1][j] + al);
      dp[i][j] = max(dp[i][j], dp[i - 1][3] + al);
    }
  }
  printf("%lld\n", dp[n][2]);
}
int main() {
  build();
  while (__--) {
    init();
    solve();
  }
}
