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 = 1009;
const long long oo = 1000;
const double LOG_2_3 = log(3) / log(2);
const long long module = (long long)(1e9) + 7ll;
int n;
int a[N][N];
int dp[5][N][N];
int cou[5][5][N][N];
pair<int, int> res;
bool ok;
long long cal(pair<int, int> kk) {
  if (!ok) return 0;
  long long ans = 1;
  for (int i = 1; i <= kk.first; i++) ans *= 2, ans %= module;
  for (int i = 1; i <= kk.second; i++) ans *= 3, ans %= module;
  return ans;
}
void update(int k1, int k2) {
  double log2res = (double)res.first + (double)res.second * LOG_2_3;
  double log2upd = (double)k1 + (double)k2 * LOG_2_3;
  if (log2upd > log2res) res = make_pair(k1, k2);
}
int main() {
  ios::sync_with_stdio(false);
  res = make_pair(0, 0);
  ok = 0;
  cin >> n;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      char x;
      cin >> x;
      a[i][j] = x - '0';
      if (x != '0') ok = 1;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      dp[1][i][j] = dp[2][i][j] = 0;
      if (a[i][j]) {
        dp[1][i][j] = dp[1][i - 1][j] + 1;
        cou[1][2][i][j] = cou[1][2][i - 1][j] + (a[i][j] == 2);
        cou[1][3][i][j] = cou[1][3][i - 1][j] + (a[i][j] == 3);
        dp[2][i][j] = dp[2][i][j - 1] + 1;
        cou[2][2][i][j] = cou[2][2][i][j - 1] + (a[i][j] == 2);
        cou[2][3][i][j] = cou[2][3][i][j - 1] + (a[i][j] == 3);
      }
    }
  for (int i = n; i >= 1; i--)
    for (int j = n; j >= 1; j--) {
      dp[3][i][j] = dp[4][i][j] = 0;
      if (a[i][j]) {
        dp[3][i][j] = dp[3][i][j + 1] + 1;
        cou[3][2][i][j] = cou[3][2][i][j + 1] + (a[i][j] == 2);
        cou[3][3][i][j] = cou[3][3][i][j + 1] + (a[i][j] == 3);
        dp[4][i][j] = dp[4][i + 1][j] + 1;
        cou[4][2][i][j] = cou[4][2][i + 1][j] + (a[i][j] == 2);
        cou[4][3][i][j] = cou[4][3][i + 1][j] + (a[i][j] == 3);
      }
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int k = min(min(dp[1][i][j], dp[2][i][j]), min(dp[3][i][j], dp[4][i][j]));
      int cnt2 = cou[1][2][i][j] - cou[1][2][i - k][j] + cou[2][2][i][j] -
                 cou[2][2][i][j - k] + cou[3][2][i][j] - cou[3][2][i][j + k] +
                 cou[4][2][i][j] - cou[4][2][i + k][j];
      int cnt3 = cou[1][3][i][j] - cou[1][3][i - k][j] + cou[2][3][i][j] -
                 cou[2][3][i][j - k] + cou[3][3][i][j] - cou[3][3][i][j + k] +
                 cou[4][3][i][j] - cou[4][3][i + k][j];
      cnt2 -= (a[i][j] == 2) * 3;
      cnt3 -= (a[i][j] == 3) * 3;
      update(cnt2, cnt3);
    }
  memset(dp, 0, sizeof dp);
  memset(cou, 0, sizeof cou);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      dp[1][i][j] = 0;
      if (a[i][j]) {
        dp[1][i][j] = dp[1][i - 1][j - 1] + 1;
        cou[1][2][i][j] = cou[1][2][i - 1][j - 1] + (a[i][j] == 2);
        cou[1][3][i][j] = cou[1][3][i - 1][j - 1] + (a[i][j] == 3);
      }
    }
    for (int j = n; j >= 1; j--) {
      dp[2][i][j] = 0;
      if (a[i][j]) {
        dp[2][i][j] = dp[2][i - 1][j + 1] + 1;
        cou[2][2][i][j] = cou[2][2][i - 1][j + 1] + (a[i][j] == 2);
        cou[2][3][i][j] = cou[2][3][i - 1][j + 1] + (a[i][j] == 3);
      }
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 1; j <= n; j++) {
      dp[3][i][j] = 0;
      if (a[i][j]) {
        dp[3][i][j] = dp[3][i + 1][j - 1] + 1;
        cou[3][2][i][j] = cou[3][2][i + 1][j - 1] + (a[i][j] == 2);
        cou[3][3][i][j] = cou[3][3][i + 1][j - 1] + (a[i][j] == 3);
      }
      for (int j = n; j >= 1; j--) {
        dp[4][i][j] = 0;
        if (a[i][j]) {
          dp[4][i][j] = dp[4][i + 1][j + 1] + 1;
          cou[4][2][i][j] = cou[4][2][i + 1][j + 1] + (a[i][j] == 2);
          cou[4][3][i][j] = cou[4][3][i + 1][j + 1] + (a[i][j] == 3);
        }
      }
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int k = min(min(dp[1][i][j], dp[2][i][j]), min(dp[3][i][j], dp[4][i][j]));
      int cnt2 = cou[1][2][i][j] - cou[1][2][i - k][j - k] + cou[2][2][i][j] -
                 cou[2][2][i - k][j + k] + cou[3][2][i][j] -
                 cou[3][2][i + k][j - k] + cou[4][2][i][j] -
                 cou[4][2][i + k][j + k];
      int cnt3 = cou[1][3][i][j] - cou[1][3][i - k][j - k] + cou[2][3][i][j] -
                 cou[2][3][i - k][j + k] + cou[3][3][i][j] -
                 cou[3][3][i + k][j - k] + cou[4][3][i][j] -
                 cou[4][3][i + k][j + k];
      cnt2 -= (a[i][j] == 2) * 3;
      cnt3 -= (a[i][j] == 3) * 3;
      update(cnt2, cnt3);
    }
  cout << cal(res);
}
