Matrix of given integers


a1,1 a1,2 ... a1, n
a2,1 a2,2 ... a2, n
::
an, 1 an, 2 ... an, n


Then, create a program that outputs the maximum value of the sum of one or more consecutive terms (submatrix) in the vertical and horizontal directions and ends.



Input

The input data is given in the following format.


n
a1,1 a1,2 ... a1, n
a2,1 a2,2 ... a2, n
::
an, 1 an, 2 ... an, n


n is 1 or more and 100 or less, and ai, j is -10000 or more and 10000 or less.

Output

Print the maximum value on one line.

Examples

Input

3
1 -2 3
-4 5 6
7 8 -9


Output

16


Input

4
1 3 -9 2
2 7 -1 5
-8 3 2 -1
5 0 -3 1


Output

15
#include <iostream>
#include <iomanip>
#include <cassert>
#include <algorithm>
#include <functional>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <bitset>
#include <sstream>
#include <istream>
#include <cmath>
#include <cstdio>
#include <complex>

using namespace std;

#define vci vector<int>
#define vcs vector<string>
#define pb push_back
#define sz size()
#define mapii map<int, int>
#define mapci map<char, int>
#define mapsi map<string, int>
#define all(x) x.begin(), x.end()
#define minit(a, i) memset(a, i, sizeof(a));

#define for_(i, a, b) for (int i=(int)a; i<(int)b; i++)
#define for_d(i, a, b) for (int i=(int)a-1; i>=b; i--)
#define for_r(i, a, b, c) for (int i=(int)a; i<(int)b; i += c)
#define for_dr(i, a, b, c) for (int i=(int)a-1; i>=b; i -= c)
#define FOR(i,c) for(__typeof((c).begin())i=(c).begin();i!=(c).end();++i)

template <class T> int __builtin_popcount(T n) { return n ? 1 + __builtin_popcount(n & (n - 1)) : 0; }

typedef long long ll;
typedef double D;

const int iINF = 2147483647;
const ll lINF = 9223372036854775807;

template <class T> inline void dbg(T t) { cout << t << endl; }




int main() {
	int n; cin >> n;
	
	int v[110];
	int b[110][110]; minit(b, -1);
	int ans = -1000000000;
	
	for_(i, 0, n) {
		for_(j, 0, n) cin >> v[j];
		
		for_(j, 0, n) {
			int s = 0;
			for_(k, j, n) {
				s += v[k];
				b[j][k] = max(b[j][k], 0) + s;
				ans = max(ans, b[j][k]);
			}
		}
	}
	
	cout << ans << endl;
	
	return 0;
}