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;
int dx[8] = {0, -1, -1, -1, 0, 1, 1, 1};
int dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
int n, p[8][1005][1005], a[1005][1005], num, x, y, ans;
char c[1005][1005];
double lg[1005][1005], sum[8][1005][1005], res = -1;
const int mod = 1e9 + 7;
bool flag1;
void pre1() {
  double lg2 = log(2);
  double lg3 = log(3);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      cin >> c[i][j];
      a[i][j] = c[i][j] - '0';
      if (a[i][j] == 3)
        lg[i][j] = lg3;
      else if (a[i][j] == 2)
        lg[i][j] = lg2;
      if (a[i][j] != 0) {
        for (int k = 0; k < 4; k++) p[k][i][j] = p[k][i + dx[k]][j + dy[k]] + 1;
        ans = 1;
      }
    }
  for (int i = n; i >= 1; i--)
    for (int j = n; j >= 1; j--)
      if (a[i][j] != 0)
        for (int k = 4; k < 8; k++) p[k][i][j] = p[k][i + dx[k]][j + dy[k]] + 1;
}
void pre2() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      for (int k = 0; k < 4; k++)
        sum[k][i][j] = sum[k][i + dx[k]][j + dy[k]] + lg[i][j];
  for (int i = n; i >= 1; i--)
    for (int j = n; j >= 1; j--)
      for (int k = 4; k < 8; k++)
        sum[k][i][j] = sum[k][i + dx[k]][j + dy[k]] + lg[i][j];
}
void solve() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (a[i][j] != 0) {
        int tmp = 1e5;
        double tmp1 = 0;
        for (int k = 1; k < 8; k += 2) tmp = min(tmp, p[k][i][j]);
        for (int k = 1; k < 8; k += 2)
          tmp1 += sum[k][i][j] - sum[k][i + tmp * dx[k]][j + tmp * dy[k]];
        tmp1 -= 3 * lg[i][j];
        if (tmp1 > res) {
          flag1 = 1;
          num = tmp;
          res = tmp1;
          x = i;
          y = j;
        }
        tmp = 1e5;
        tmp1 = 0;
        for (int k = 0; k < 7; k += 2) tmp = min(tmp, p[k][i][j]);
        for (int k = 0; k < 7; k += 2)
          tmp1 += sum[k][i][j] - sum[k][i + tmp * dx[k]][j + tmp * dy[k]];
        tmp1 -= 3 * lg[i][j];
        if (tmp1 > res) {
          flag1 = 0;
          num = tmp;
          res = tmp1;
          x = i;
          y = j;
        }
      }
  if (flag1) {
    for (int i = 1; i < num; i++)
      for (int k = 1; k < 8; k += 2)
        ans = (1LL * ans * a[x + i * dx[k]][y + i * dy[k]]) % mod;
    ans = (1LL * ans * a[x][y]) % mod;
  } else {
    for (int i = 1; i < num; i++)
      for (int k = 0; k < 8; k += 2)
        ans = (1LL * ans * a[x + i * dx[k]][y + i * dy[k]]) % mod;
    ans = (1LL * ans * a[x][y]) % mod;
  }
  cout << ans << endl;
}
int main() {
  cin >> n;
  pre1();
  pre2();
  solve();
}
