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 = 1005;
const int mod = 1e9 + 7;
char str[N];
int arr[N][N];
int n;
int cnt1[N][N][4];
int cnt2[N][N][4];
int cnt3[N][N][4];
int cnt4[N][N][4];
vector<pair<int, int> > ans;
const long double log3 = 1.58496250072115618145373894395;
bool cmp(pair<int, int> a, pair<int, int> b) {
  return (a.first + log3 * a.second > b.first + log3 * b.second);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", str);
    for (int j = 0; j < n; ++j) {
      arr[i][j + 1] = str[j] - '0';
    }
  }
  memset(cnt1, 0, sizeof(cnt1));
  memset(cnt2, 0, sizeof(cnt2));
  memset(cnt3, 0, sizeof(cnt3));
  memset(cnt4, 0, sizeof(cnt4));
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k < 4; ++k) {
        cnt1[i][j][k] = cnt1[i][j - 1][k];
      }
      ++cnt1[i][j][arr[i][j]];
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k < 4; ++k) {
        cnt2[i][j][k] = cnt2[i - 1][j][k];
      }
      ++cnt2[i][j][arr[i][j]];
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k < 4; ++k) {
        cnt3[i][j][k] = cnt3[i - 1][j - 1][k];
      }
      ++cnt3[i][j][arr[i][j]];
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      for (int k = 0; k < 4; ++k) {
        cnt4[i][j][k] = cnt4[i - 1][j + 1][k];
      }
      ++cnt4[i][j][arr[i][j]];
    }
  }
  ans.clear();
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (arr[i][j]) {
        int sz = min(i, n - i + 1);
        sz = min(sz, min(j, n - j + 1));
        int l = 2;
        int r = sz + 1;
        while (l < r) {
          int mid = l + r >> 1;
          int x, y;
          y = j;
          x = i - mid + 1;
          bool rekt = 0;
          if (x < 1 || cnt2[i][j][0] - cnt2[x - 1][j][0]) {
            rekt = 1;
          }
          x = i + mid - 1;
          if (x > n || cnt2[x][j][0] - cnt2[i - 1][j][0]) {
            rekt = 1;
          }
          x = i;
          y = j + mid - 1;
          if (y > n || cnt1[x][y][0] - cnt1[i][j - 1][0]) {
            rekt = 1;
          }
          y = j - mid + 1;
          if (y < 1 || cnt1[i][j][0] - cnt1[i][y - 1][0]) {
            rekt = 1;
          }
          if (rekt) {
            r = mid;
          } else {
            l = mid + 1;
          }
        }
        --l;
        --r;
        int cn2 = cnt2[i + l - 1][j][2] - cnt2[i - l][j][2];
        cn2 += cnt1[i][j + l - 1][2] - cnt1[i][j - l][2];
        cn2 -= (arr[i][j] == 2);
        int cn3 = cnt2[i + l - 1][j][3] - cnt2[i - l][j][3];
        cn3 += cnt1[i][j + l - 1][3] - cnt1[i][j - l][3];
        cn3 -= (arr[i][j] == 3);
        ans.emplace_back(make_pair(cn2, cn3));
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (arr[i][j]) {
        int sz = min(i, n - i + 1);
        sz = min(sz, min(j, n - j + 1));
        int l = 2;
        int r = sz + 1;
        while (l < r) {
          int mid = l + r >> 1;
          int x, y;
          x = i - mid + 1;
          y = j - mid + 1;
          bool rekt = 0;
          if (min(x, y) < 1 || cnt3[i][j][0] - cnt3[x - 1][y - 1][0]) {
            rekt = 1;
          }
          y = j + mid - 1;
          if (x < 1 || y > n || cnt4[i][j][0] - cnt4[x - 1][y + 1][0]) {
            rekt = 1;
          }
          x = i + mid - 1;
          y = j - mid + 1;
          if (x > n || y < 1 || cnt4[x][y][0] - cnt4[i - 1][j + 1][0]) {
            rekt = 1;
          }
          y = j + mid - 1;
          if (x > n || y > n || cnt3[x][y][0] - cnt3[i - 1][j - 1][0]) {
            rekt = 1;
          }
          if (rekt) {
            r = mid;
          } else {
            l = mid + 1;
          }
        }
        --l;
        --r;
        int cn2 = 0;
        cn2 += cnt3[i + l - 1][j + l - 1][2] - cnt3[i - l][j - l][2];
        cn2 += cnt4[i + l - 1][j - l + 1][2] - cnt4[i - l][j + l][2];
        cn2 -= (arr[i][j] == 2);
        int cn3 = 0;
        cn3 += cnt3[i + l - 1][j + l - 1][3] - cnt3[i - l][j - l][3];
        cn3 += cnt4[i + l - 1][j - l + 1][3] - cnt4[i - l][j + l][3];
        cn3 -= (arr[i][j] == 3);
        ans.emplace_back(make_pair(cn2, cn3));
      }
    }
  }
  if (ans.empty()) {
    cout << "0\n";
    return 0;
  }
  pair<int, int> print = make_pair(0, 0);
  for (auto it : ans) {
    if (cmp(it, print)) {
      print = it;
    }
  }
  int answer = 1;
  for (int i = 0; i < print.first; ++i) {
    answer = (answer * 2LL) % mod;
  }
  for (int i = 0; i < print.second; ++i) {
    answer = (answer * 3LL) % mod;
  }
  printf("%d\n", answer);
}
