/* Subject: Contest submission for problem #1, file 1.cc */
/* cs188-ah@imail.EECS.Berkeley.EDU */
/* Wed Sep 10 13:49:48 PDT 2003 */
/*___CONTEST_SUBMISSION___ cs188-ah 1 */
//Estimated time: 2hr30minute Real time: 4 hours
//Took so long to get triangle detection subroutines
//to work perfectly! Also was tricky coming up with
//surrounding matrix trick
//SID 14394533

#include <vector>
#include <set>
#include <iostream>
#include <map>
#include <string>
using namespace std;

typedef set<char>::iterator setiter;
typedef vector<string>::iterator viter;
int tri1(const vector<string>& nn, size_t n, char t, size_t x, size_t y);
int tri2(vector<string> nn, size_t n, char t, size_t x, size_t y);
int main(int argc, char* argv[])
{
  vector<string> nn(60);
  set<char> sc;
  map<char,int> c;
  size_t n, total;
  
  while (cin >> n) {
    if (n == 0)
      break;
    c.clear();
    sc.clear();
    total = 0;
    for(viter vi = nn.begin(); vi != nn.end(); vi++)
      *vi = string("000000000100000000010000000001000000000100000000010000000001");

    // surround the matrix with non letter entries, necessary to find all of tri2
    for(int ii = 20; ii < 20 + n; ii++) {
      string row;
      cin >> row;
      nn[ii].replace(20, n, row);
      //      cout << nn[ii] << endl;      
      for(int jj =0; jj < row.length(); jj++)
	sc.insert(char(row[jj]));
    }
    
    // for each matrix, find number of triangle for each kind of letter
    for(setiter si = sc.begin(); si != sc.end(); si++) {
      // look for matrices 2x2 up to to size nxn
      for(int ii = 2; ii <= n ; ii++) { 
	// start at edge of our matrix left to right
	for(int jj = 0; jj <= 60 - ii; jj++)
	  // go work our way downwards
	  for(int kk = 0; kk <= 60 - ii; kk++) {
	    c[*si] += tri1(nn, ii, *si, jj, kk);
	    c[*si] += tri2(nn, ii, *si, jj, kk);
	  }
      }
      total += c[*si];
      
    }
    cout << "(" << total << ")";
    for(setiter si = sc.begin(); si != sc.end(); si++)
      cout << " " << c[*si] << " " <<  *si;
    cout << endl;
  }
}

// find triangle of matrix n size
// only looks for triangles with legs
// alined along 2 axis 
int tri1(const vector<string>& nn, size_t n, char t, size_t x, size_t y)
{

  int t1 = 1, t2 = 1, t3 = 1, t4 = 1;

  // for 2x2 just do brute force
  if (n == 2) {
    return 
      (((nn[0 + x][0 + y] == t) && (nn[0 + x][1 + y] == t) && (nn[1 + x][0 + y] == t)) +
       ((nn[0 + x][0 + y] == t) && (nn[0 + x][1 + y] == t) && (nn[1 + x][1 + y] == t)) +
       ((nn[0 + x][0 + y] == t) && (nn[1 + x][0 + y] == t) && (nn[1 + x][1 + y] == t)) +
       ((nn[1 + x][0 + y] == t) && (nn[1 + x][1 + y] == t) && (nn[0 + x][1 + y] == t)));
  }
          
  // looks for
  // A
  // A A
  // A A A
  for(int ii = 0; ii < n; ii++)
    for(int jj = 0 ; jj < ii + 1; jj++)
      if (nn[ii + x][jj + y] != t)
	t1 = 0;

  // looks for
  // A A A
  // A A
  // A 
  for(int ii = 0; ii < n; ii++)
    for(int jj = 0; jj < n - ii; jj++)
      if (nn[ii + x][jj + y] != t)
	t2 = 0;

  // looks for
  // A A A
  //   A A
  //     A 
  for(int ii = 0; ii < n; ii++)
    for(int jj = ii; jj < n; jj++)
      if (nn[ii + x][jj + y] != t)
	t3 = 0;

  // looks for
  //    A
  //  A A
  //A A A
  for(int ii = 0; ii < n; ii++)
    for(int jj = n - ii - 1; jj < n; jj++)
      if (nn[ii + x][jj + y] != t)
	t4 = 0;

  return t1 + t2 + t3 + t4;
}
	
int tri2(vector<string> nn, size_t n, char t, size_t x, size_t y)
{

  int t1 = 1, t2 = 1, t3 = 1, t4 = 1;
  int np = n + 1;
  // only possible in odd size matrix
  if ((n % 2) == 0) return 0;

  // look for  
  // A A A A A
  //   A A A 
  //     A 
  for(int ii = 0, kk = n; ii < (np/2); ii++, kk--)
    for(int jj = ii; jj < kk; jj++)
      if (nn[ii + x][jj + y] != t)
	t1 = 0;

  // look for 
  //     A
  //   A A A
  // A A A A A  
  for(int ii = (np/2) - 1, kk = 0; ii < n; ii++, kk++)
    for(int jj = (np/2) - kk -1; jj < (np/2) + kk ; jj++)
      if (nn[ii + x][jj + y] != t)
	t2 = 0; 

  // look for
  // A 
  // A A 
  // A A A
  // A A
  // A
  for(int ii = 0, kk = n; ii < n; ii++,kk--) {
    int zz = (ii > kk) ? kk : ii + 1;
    for(int jj = 0; jj < zz; jj++) {
      if (nn[ii + x][jj + y] != t)
	t3 = 0;
    }
  }
  // look for
  //      A
  //    A A
  //  A A A
  //    A A
  //      A
  for(int ii = 0, kk = n - 1; ii < n; ii++, kk--) {
    int zz = (ii > kk) ? ii : kk;
    for(int jj = zz; jj < n ; jj++)
      if (nn[ii + x][jj + y] != t)
	t4 = 0; 
  }

  return t1 + t2 + t3 + t4;
}