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;
template <typename T>
T next_int() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') {
    p = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * p;
}
const int max_n = (int)1e3 + 227 + 1;
const int max_int = (int)1e9 + 227 + 1;
const int mod = (int)1e9 + 7;
int dp[4][max_n][max_n][4];
int main() {
  int n = next_int<int>();
  vector<string> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    a[i] = '?' + a[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      for (int k = 0; k <= 3; k++) {
        dp[0][i][j][k] = dp[0][i - 1][j][k];
        dp[1][i][j][k] = dp[1][i][j - 1][k];
        if (a[i][j] - '0' == k) dp[1][i][j][k]++;
        dp[2][i][j][k] = dp[2][i - 1][j - 1][k];
      }
    }
    for (int j = n; j >= 1; j--) {
      for (int k = 0; k <= 3; k++) {
        dp[3][i][j][k] = dp[3][i - 1][j + 1][k];
      }
      dp[0][i][j][a[i][j] - '0']++;
      dp[2][i][j][a[i][j] - '0']++;
      dp[3][i][j][a[i][j] - '0']++;
    }
  }
  pair<long double, pair<int, int> > ans = make_pair(-1, make_pair(0, 0));
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i][j] == '0') continue;
      int l = 0;
      int r = min(min(i - 1, j - 1), min(n - i, n - j));
      while (r - l > 1) {
        int c = (l + r) / 2;
        if ((dp[0][i + c][j][0] - dp[0][i - c - 1][j][0]) +
                (dp[1][i][j + c][0] - dp[1][i][j - c - 1][0]) ==
            0)
          l = c;
        else
          r = c;
      }
      int c = r;
      if ((dp[0][i + c][j][0] - dp[0][i - c - 1][j][0]) +
          (dp[1][i][j + c][0] - dp[1][i][j - c - 1][0]))
        c = l;
      int k[2];
      k[0] = (dp[0][i + c][j][2] - dp[0][i - c - 1][j][2]) +
             (dp[1][i][j + c][2] - dp[1][i][j - c - 1][2]);
      k[1] = (dp[0][i + c][j][3] - dp[0][i - c - 1][j][3]) +
             (dp[1][i][j + c][3] - dp[1][i][j - c - 1][3]);
      if (a[i][j] == '2') k[0]--;
      if (a[i][j] == '3') k[1]--;
      if (log10(2) * k[0] + log10(3) * k[1] > ans.first) {
        ans.first = log10(2) * k[0] + log10(3) * k[1];
        ans.second.first = k[0];
        ans.second.second = k[1];
      }
      l = 0;
      r = min(min(i - 1, j - 1), min(n - i, n - j));
      while (r - l > 1) {
        int c = (l + r) / 2;
        if ((dp[2][i + c][j + c][0] - dp[2][i - c - 1][j - c - 1][0]) +
                (dp[3][i + c][j - c][0] - dp[3][i - c - 1][j + c + 1][0]) ==
            0)
          l = c;
        else
          r = c;
      }
      c = r;
      if ((dp[2][i + c][j + c][0] - dp[2][i - c - 1][j - c - 1][0]) +
          (dp[3][i + c][j - c][0] - dp[3][i - c - 1][j + c + 1][0]))
        c = l;
      k[0] = (dp[2][i + c][j + c][2] - dp[2][i - c - 1][j - c - 1][2]) +
             (dp[3][i + c][j - c][2] - dp[3][i - c - 1][j + c + 1][2]);
      k[1] = (dp[2][i + c][j + c][3] - dp[2][i - c - 1][j - c - 1][3]) +
             (dp[3][i + c][j - c][3] - dp[3][i - c - 1][j + c + 1][3]);
      if (a[i][j] == '2') k[0]--;
      if (a[i][j] == '3') k[1]--;
      if (log10(2) * k[0] + log10(3) * k[1] > ans.first) {
        ans.first = log10(2) * k[0] + log10(3) * k[1];
        ans.second.first = k[0];
        ans.second.second = k[1];
      }
    }
  }
  long long mda = 1;
  if (ans.first == -1) mda = 0;
  while (ans.second.first) {
    mda = mda * 2 % mod;
    ans.second.first--;
  }
  while (ans.second.second) {
    mda = mda * 3 % mod;
    ans.second.second--;
  }
  cout << mda << "\n";
}
