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>
#pragma warning(disable : 4996)
using namespace std;
const int maxn = 1005;
const long long mod = 1e9 + 7;
const double PI = acos(-1.0);
int dx[8] = {0, -1, -1, -1, 1, 1, 0, 1};
int dy[8] = {-1, -1, 0, 1, 0, -1, 1, 1};
int n;
char tmp[maxn];
int val[maxn][maxn];
double logv[maxn][maxn];
int len[8][maxn][maxn];
double dp[8][maxn][maxn];
void solve() {
  int i, j, k;
  double lg2 = log(2.0);
  double lg3 = log(3.0);
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    scanf("%s", tmp + 1);
    for (int j = 1; j <= n; j++) {
      val[i][j] = tmp[j] - '0';
      if (val[i][j] == 2) {
        logv[i][j] = lg2;
      } else if (val[i][j] == 3) {
        logv[i][j] = lg3;
      }
    }
  }
  memset(len, 0, sizeof(len));
  memset(dp, 0, sizeof(dp));
  int dir;
  for (dir = 0; dir < 4; dir++) {
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        if (val[i][j]) {
          int nx = i + dx[dir];
          int ny = j + dy[dir];
          len[dir][i][j] = len[dir][nx][ny] + 1;
          dp[dir][i][j] = dp[dir][nx][ny] + logv[i][j];
        }
      }
    }
  }
  for (dir = 4; dir < 8; dir++) {
    for (i = n; i >= 1; i--) {
      for (j = n; j >= 1; j--) {
        if (val[i][j]) {
          int nx = i + dx[dir];
          int ny = j + dy[dir];
          len[dir][i][j] = len[dir][nx][ny] + 1;
          dp[dir][i][j] = dp[dir][nx][ny] + logv[i][j];
        }
      }
    }
  }
  double res = -1;
  int resx = 0, resy = 0, reslen = 0, d = 0;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (val[i][j]) {
        int tmp = n + 5;
        for (k = 0; k < 8; k += 2) {
          tmp = min(tmp, len[k][i][j]);
        }
        double ans = 0;
        for (k = 0; k < 8; k += 2) {
          int nx = i + (tmp)*dx[k];
          int ny = j + (tmp)*dy[k];
          ans += dp[k][i][j] - dp[k][nx][ny];
        }
        if (val[i][j] == 2) {
          ans -= 3 * lg2;
        } else if (val[i][j] == 3) {
          ans -= 3 * lg3;
        }
        if (ans > res) {
          res = ans;
          resx = i;
          resy = j;
          reslen = tmp;
          d = 0;
        }
        tmp = n + 5;
        for (k = 1; k < 8; k += 2) {
          tmp = min(tmp, len[k][i][j]);
        }
        ans = 0;
        for (k = 1; k < 8; k += 2) {
          int nx = i + (tmp)*dx[k];
          int ny = j + (tmp)*dy[k];
          ans += dp[k][i][j] - dp[k][nx][ny];
        }
        if (val[i][j] == 2) {
          ans -= 3 * lg2;
        } else if (val[i][j] == 3) {
          ans -= 3 * lg3;
        }
        if (ans > res) {
          res = ans;
          resx = i;
          resy = j;
          reslen = tmp;
          d = 1;
        }
      }
    }
  }
  if (res < 0) {
    printf("0\n");
  } else {
    long long r = 1;
    for (k = d; k < 8; k += 2) {
      for (j = 2; j <= reslen; j++) {
        int nx = resx + (j - 1) * dx[k];
        int ny = resy + (j - 1) * dy[k];
        r *= val[nx][ny];
        r %= mod;
      }
    }
    r *= val[resx][resy];
    r %= mod;
    printf("%lld\n", r);
  }
}
int main() {
  solve();
  return 0;
}
