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 di[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dj[] = {0, 1, 1, 1, 0, -1, -1, -1};
int mark[1010][1010][8];
int dist[1010][1010][8];
int pre[1010][1010][8][4];
char s[1010][1010];
long long p[4][10000];
double dp[2010][2010];
void go(int i, int j, int d) {
  if (mark[i][j][d]) return;
  mark[i][j][d] = 1;
  go(i + di[d], j + dj[d], d);
  if (s[i - 1][j - 1] == '0')
    dist[i][j][d] = 0;
  else
    dist[i][j][d] = 1 + dist[i + di[d]][j + dj[d]][d];
  for (int k = 0; k < 4; ++k) pre[i][j][d][k] = pre[i + di[d]][j + dj[d]][d][k];
  pre[i][j][d][s[i - 1][j - 1] - '0']++;
}
int main() {
  int n, i, j, k, d;
  bool f = 0;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) {
    scanf("%s", &s[i]);
    for (j = 0; j < n; ++j)
      if (s[i][j] != '0') f = 1;
  }
  if (!f) return 0 * printf("0\n");
  for (i = 0; i < 1010; ++i)
    for (j = 0; j < 1010; ++j)
      for (d = 0; d < 8; ++d) {
        mark[i][j][d] = dist[i][j][d] = 0;
        for (k = 0; k < 4; ++k) pre[i][j][d][k] = 0;
      }
  for (i = 0; i <= n + 1; ++i)
    for (d = 0; d < 8; ++d)
      mark[0][i][d] = mark[n + 1][i][d] = mark[i][0][d] = mark[i][n + 1][d] = 1;
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      for (d = 0; d < 8; ++d) go(i, j, d);
  for (i = 2; i <= 3; ++i) {
    p[i][0] = 1;
    for (j = 1; j < 10000; ++j) p[i][j] = (p[i][j - 1] * i) % 1000000007;
  }
  dp[0][0] = 0;
  for (i = 1; i <= 2000; ++i) dp[0][i] = dp[0][i - 1] + log10(3.0);
  for (i = 1; i <= 2000; ++i)
    for (j = 0; j <= 2000; ++j) dp[i][j] = dp[i - 1][j] + log10(2.0);
  int chi, chj;
  pair<double, pair<int, int> > maxx = make_pair(0.0, make_pair(0, 0));
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      for (int st = 0; st <= 1; ++st) {
        int minn = 10000;
        for (d = st; d < 8; d += 2) minn = min(minn, dist[i][j][d]);
        int cnt[4];
        cnt[2] = cnt[3] = 0;
        for (d = st; d < 8; d += 2)
          for (k = 2; k <= 3; ++k)
            cnt[k] += (pre[i][j][d][k] -
                       pre[i + di[d] * minn][j + dj[d] * minn][d][k]);
        if (s[i - 1][j - 1] == '2') cnt[2] -= 3;
        if (s[i - 1][j - 1] == '3') cnt[3] -= 3;
        maxx =
            max(maxx, make_pair(dp[cnt[2]][cnt[3]], make_pair(cnt[2], cnt[3])));
      }
  printf("%lld\n",
         (p[2][maxx.second.first] * p[3][maxx.second.second]) % 1000000007);
}
