Vanya plays a game of balloons on the field of size n × n, where each cell contains a balloon with one of the values 0, 1, 2 or 3. The goal is to destroy a cross, such that the product of all values of balloons in the cross is maximum possible. There are two types of crosses: normal and rotated. For example:
    
    
      
    **o**  
    **o**  
    ooooo  
    **o**  
    **o**  
    

or
    
    
      
    o***o  
    *o*o*  
    **o**  
    *o*o*  
    o***o  
    

Formally, the cross is given by three integers r, c and d, such that d ≤ r, c ≤ n - d + 1. The normal cross consists of balloons located in cells (x, y) (where x stay for the number of the row and y for the number of the column), such that |x - r|·|y - c| = 0 and |x - r| + |y - c| < d. Rotated cross consists of balloons located in cells (x, y), such that |x - r| = |y - c| and |x - r| < d.

Vanya wants to know the maximum possible product of the values of balls forming one cross. As this value can be large, output it modulo 109 + 7.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of rows and columns in the table with balloons.

The each of the following n lines contains n characters '0', '1', '2' or '3' — the description of the values in balloons.

Output

Print the maximum possible product modulo 109 + 7. Note, that you are not asked to maximize the remainder modulo 109 + 7, but to find the maximum value and print it this modulo.

Examples

Input

4
1233
0213
2020
0303


Output

108


Input

5
00300
00300
33333
00300
00300


Output

19683


Input

5
00003
02030
00300
03020
30000


Output

108


Input

5
21312
10003
10002
10003
23231


Output

3


Input

5
12131
12111
12112
21311
21212


Output

24

Note

In the first sample, the maximum product is achieved for a rotated cross with a center in the cell (3, 3) and radius 1: 2·2·3·3·3 = 108.
#include <bits/stdc++.h>
using namespace std;
const int N = 2e3 + 7;
const int MOD = 1e9 + 7;
int a[N][N];
long double R[N][N], C[N][N];
int d[N][N];
bool f[N][N];
string b[N];
int n;
pair<pair<long double, int>, pair<int, int> > solve() {
  memset(R, 0, sizeof R);
  memset(C, 0, sizeof C);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      if (f[i][j] && a[i][j]) {
        R[i][j + 1] += log(a[i][j]);
        C[i + 1][j] += log(a[i][j]);
      }
  for (int i = 0; i < N; i++)
    for (int j = 0; j + 1 < N; j++) R[i][j + 1] += R[i][j];
  for (int i = 0; i + 1 < N; i++)
    for (int j = 0; j < N; j++) C[i + 1][j] += C[i][j];
  for (int i = 0; i < N; i++) {
    int prv = -2 * N;
    for (int j = 0; j < N; j++) {
      if (f[i][j] && a[i][j] == 0) prv = j;
      d[i][j] = min(d[i][j], j - prv);
    }
    prv = 2 * N;
    for (int j = N - 1; j >= 0; j--) {
      if (f[i][j] && a[i][j] == 0) prv = j;
      d[i][j] = min(d[i][j], prv - j);
    }
  }
  for (int j = 0; j < N; j++) {
    int prv = -2 * N;
    for (int i = 0; i < N; i++) {
      if (f[i][j] && a[i][j] == 0) prv = i;
      d[i][j] = min(d[i][j], i - prv);
    }
    prv = 2 * N;
    for (int i = N - 1; i >= 0; i--) {
      if (f[i][j] && a[i][j] == 0) prv = i;
      d[i][j] = min(d[i][j], prv - i);
    }
  }
  long double sum = -1;
  int rr = n - 1, cc = 0;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) {
      if (f[i][j] && d[i][j]) {
      }
      if (f[i][j] && d[i][j] &&
          sum < R[i][min(N - 1, j + d[i][j])] - R[i][max(0, j - d[i][j] + 1)] +
                    C[min(N - 1, i + d[i][j])][j] -
                    C[max(0, i - d[i][j] + 1)][j] - log(a[i][j])) {
        sum = R[i][min(N - 1, j + d[i][j])] - R[i][max(0, j - d[i][j] + 1)] +
              C[min(N - 1, i + d[i][j])][j] - C[max(0, i - d[i][j] + 1)][j] -
              log(a[i][j]);
        rr = i, cc = j;
      }
    }
  int ans = 1;
  for (int i = 0; i < N; i++)
    if (f[i][cc] && i != rr && abs(i - rr) < d[rr][cc])
      ans = 1LL * ans * a[i][cc] % MOD;
  for (int i = 0; i < N; i++)
    if (f[rr][i] && i != cc && abs(i - cc) < d[rr][cc])
      ans = 1LL * ans * a[rr][i] % MOD;
  ans = 1LL * ans * a[rr][cc] % MOD;
  return {{sum, ans}, {rr, cc}};
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> b[i];
  memset(d, 63, sizeof d);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) a[i][j] = 1, f[i][j] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      f[i][j] = 1;
      a[i][j] = b[i][j] - '0';
      d[i][j] = min(min(i + 1, j + 1), min(n - i, n - j));
    }
  auto p = solve();
  memset(d, 63, sizeof d);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++) a[i][j] = 1, f[i][j] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      f[i - j + n - 1][i + j] = 1;
      d[i - j + n - 1][i + j] =
          2 * min(min(i + 1, j + 1), min(n - i, n - j)) - 1;
      a[i - j + n - 1][i + j] = b[i][j] - '0';
    }
  auto q = solve();
  if (p.first.first <= q.first.first)
    cout << q.first.second << "\n";
  else
    cout << p.first.second << "\n";
  return 0;
}
