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;
char a[1011][1011];
int n, m;
int dx[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, -1, 1, -1, 1};
pair<int, int> dp[1011][1011][8];
bool vis[1011][1011][8];
pair<int, int> rec(int i, int j, int d) {
  if (i > n or i < 1 or j < 1 or j > m) return make_pair(0, 0);
  if (vis[i][j][d]) return dp[i][j][d];
  vis[i][j][d] = 1;
  pair<int, int> p = rec(i + dx[d], j + dy[d], d);
  if (a[i][j] == '2') p.first++;
  if (a[i][j] == '3') p.second++;
  return dp[i][j][d] = p;
}
int sdp[1011][1011][8] = {0};
bool vis1[1011][1011][8] = {0};
int rec1(int i, int j, int d) {
  if (i > n or i < 1 or j < 1 or j > m) return 0;
  if (vis1[i][j][d]) return sdp[i][j][d];
  vis1[i][j][d] = 1;
  int p = rec1(i + dx[d], j + dy[d], d);
  if (a[i][j] == '0')
    p = 0;
  else
    p++;
  return sdp[i][j][d] = p;
}
int main() {
  cin >> n;
  m = n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  }
  double ans = -1;
  pair<int, int> need;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (a[i][j] == '0') continue;
      int d = INT_MAX;
      for (int second = 0; second < 4; second++) {
        d = min(d, rec1(i, j, second));
      }
      pair<int, int> cur = make_pair(0, 0);
      for (int second = 0; second < 4; second++) {
        pair<int, int> p = rec(i, j, second);
        pair<int, int> q = rec(i + dx[second] * d, j + dy[second] * d, second);
        cur.first += p.first, cur.second += p.second;
        cur.first -= q.first, cur.second -= q.second;
      }
      cur.first -= 3 * (a[i][j] == '2');
      cur.second -= 3 * (a[i][j] == '3');
      if (ans < (double)cur.first + log2(3) * (double)cur.second) {
        ans = (double)cur.first + log2(3) * (double)cur.second;
        need = cur;
      }
      d = INT_MAX;
      for (int second = 0; second < 4; second++) {
        d = min(d, rec1(i, j, second + 4));
      }
      cur = make_pair(0, 0);
      for (int second = 4; second < 8; second++) {
        pair<int, int> p = rec(i, j, second);
        pair<int, int> q = rec(i + dx[second] * d, j + dy[second] * d, second);
        cur.first += p.first, cur.second += p.second;
        cur.first -= q.first, cur.second -= q.second;
      }
      cur.first -= 3 * (a[i][j] == '2');
      cur.second -= 3 * (a[i][j] == '3');
      if (ans < (double)cur.first + log2(3) * (double)cur.second) {
        ans = (double)cur.first + log2(3) * (double)cur.second;
        need = cur;
      }
    }
  }
  if (ans == -1) {
    cout << 0;
    return 0;
  }
  long long pt = 1;
  for (int i = 0; i < need.first; i++) pt *= 2, pt %= 1000000007;
  for (int j = 0; j < need.second; j++) pt *= 3, pt %= 1000000007;
  cout << pt;
}
