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 n;
char str[1010][1010];
int asdf[2010][2010];
const double INF = 1000000;
const int mod = 1e9 + 7;
double ans = -INF;
int ansm = 0;
double myLog(int a) {
  if (a == 0)
    return -INF;
  else
    return log(a);
}
double A[2010][2010], B[2010][2010];
int Am[2010][2010], Bm[2010][2010];
int rAm[2010][2010], rBm[2010][2010];
int rr[2010][2010], a[2010][2010], b[2010][2010], c[2010][2010], d[2010][2010],
    near[2010][2010];
void check1() {
  for (int i = 1; i <= n; i++) Am[i][0] = Bm[0][i] = rAm[i][0] = rBm[0][i] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int tn = str[i][j] - '0';
      A[i][j] = A[i][j - 1] + myLog(tn);
      Am[i][j] = Am[i][j - 1] * 1ll * max(tn, 1) % mod;
      rAm[i][j] = rAm[i][j - 1] * 1ll * ((mod + 1) / max(tn, 1)) % mod;
      B[i][j] = B[i - 1][j] + myLog(tn);
      Bm[i][j] = Bm[i - 1][j] * 1ll * max(tn, 1) % mod;
      rBm[i][j] = rBm[i - 1][j] * 1ll * ((mod + 1) / max(tn, 1)) % mod;
      rr[i][j] = (mod + 1) / max(tn, 1) % mod;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int tn = str[i][j] - '0';
      if (tn == 0)
        a[i][j] = 0;
      else
        a[i][j] = a[i][j - 1] + 1;
      if (tn == 0)
        c[i][j] = 0;
      else
        c[i][j] = c[i - 1][j] + 1;
    }
  for (int i = n; i >= 1; i--)
    for (int j = n; j >= 1; j--) {
      int tn = str[i][j] - '0';
      if (tn == 0)
        b[i][j] = 0;
      else
        b[i][j] = b[i][j + 1] + 1;
      if (tn == 0)
        d[i][j] = 0;
      else
        d[i][j] = d[i + 1][j] + 1;
      near[i][j] = min({a[i][j], b[i][j], c[i][j], d[i][j]});
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      int mid = min({i, n - i + 1, j, n - j + 1, near[i][j]});
      if (mid == 0) continue;
      double res = A[i][j + mid - 1] - A[i][j - mid] + B[i + mid - 1][j] -
                   B[i - mid][j] - myLog(str[i][j] - '0');
      if (ans < res) {
        ans = res;
        ansm = (Am[i][j + mid - 1] * 1ll * rAm[i][j - mid] % mod) * 1ll *
               (Bm[i + mid - 1][j] * 1ll * rBm[i - mid][j] % mod) % mod;
        ansm = ansm * 1ll * rr[i][j] % mod;
      }
    }
}
void check2() {
  for (int i = 1; i <= n; i++)
    for (int j = n - i + 1; j <= 2 * n - (n - i + 1); j++) asdf[i][j] = 1;
  for (int i = 2 * n - 1; i >= n; i--)
    for (int j = i - n + 1; j <= 2 * n - (i - n + 1); j++) asdf[i][j] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = n - i + 1, k = 1; k <= i; j += 2, k++)
      asdf[i][j] = str[i - k + 1][k] - '0';
  for (int i = 2 * n - 1; i >= n; i--)
    for (int j = i - n + 1, k = 1; k <= 2 * n - i; j += 2, k++)
      asdf[i][j] = str[n - k + 1][i - n + k] - '0';
  for (int i = 1; i <= 2 * n; i++)
    Am[i][0] = Bm[0][i] = rAm[i][0] = rBm[0][i] = 1;
  for (int i = 1; i <= 2 * n; i++)
    for (int j = 1; j <= 2 * n; j++) {
      int tn = asdf[i][j];
      A[i][j] = A[i][j - 1] + myLog(tn);
      Am[i][j] = Am[i][j - 1] * 1ll * max(tn, 1) % mod;
      rAm[i][j] = rAm[i][j - 1] * 1ll * ((mod + 1) / max(tn, 1)) % mod;
      B[i][j] = B[i - 1][j] + myLog(tn);
      Bm[i][j] = Bm[i - 1][j] * 1ll * max(tn, 1) % mod;
      rBm[i][j] = rBm[i - 1][j] * 1ll * ((mod + 1) / max(tn, 1)) % mod;
      rr[i][j] = (mod + 1) / max(tn, 1) % mod;
    }
  for (int i = 1; i <= 2 * n; i++)
    for (int j = 1; j <= 2 * n; j++) {
      int tn = asdf[i][j];
      if (tn == 0)
        a[i][j] = 0;
      else
        a[i][j] = a[i][j - 1] + 1;
      if (tn == 0)
        c[i][j] = 0;
      else
        c[i][j] = c[i - 1][j] + 1;
    }
  for (int i = 2 * n; i >= 1; i--)
    for (int j = 2 * n; j >= 1; j--) {
      int tn = asdf[i][j];
      if (tn == 0)
        b[i][j] = 0;
      else
        b[i][j] = b[i][j + 1] + 1;
      if (tn == 0)
        d[i][j] = 0;
      else
        d[i][j] = d[i + 1][j] + 1;
      near[i][j] = min({a[i][j], b[i][j], c[i][j], d[i][j]});
    }
  for (int i = 1; i <= 2 * n; i++)
    for (int j = 1; j <= 2 * n; j++) {
      int mid = min({i, 2 * n - i + 1, j, 2 * n - j + 1, near[i][j]});
      if (mid == 0) continue;
      if ((1 + n) % 2 != (i + j) % 2) continue;
      double res = A[i][j + mid - 1] - A[i][j - mid] + B[i + mid - 1][j] -
                   B[i - mid][j] - myLog(asdf[i][j]);
      if (ans < res) {
        ans = res;
        ansm = (Am[i][j + mid - 1] * 1ll * rAm[i][j - mid] % mod) * 1ll *
               (Bm[i + mid - 1][j] * 1ll * rBm[i - mid][j] % mod) % mod;
        ansm = ansm * 1ll * rr[i][j] % mod;
      }
    }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%s", &(str[i][1]));
  check1();
  check2();
  printf("%d", ansm);
  return 0;
}
