Problem: You already know that Valery's favorite sport is biathlon. Due to your help, he learned to shoot without missing, and his skills are unmatched at the shooting range. But now a smaller task is to be performed, he should learn to complete the path fastest.

The track's map is represented by a rectangle n × m in size divided into squares. Each square is marked with a lowercase Latin letter (which means the type of the plot), with the exception of the starting square (it is marked with a capital Latin letters S) and the terminating square (it is marked with a capital Latin letter T). The time of movement from one square to another is equal to 1 minute. The time of movement within the cell can be neglected. We can move from the cell only to side-adjacent ones, but it is forbidden to go beyond the map edges. Also the following restriction is imposed on the path: it is not allowed to visit more than k different types of squares (squares of one type can be visited an infinite number of times). Squares marked with S and T have no type, so they are not counted. But S must be visited exactly once — at the very beginning, and T must be visited exactly once — at the very end.

Your task is to find the path from the square S to the square T that takes minimum time. Among all shortest paths you should choose the lexicographically minimal one. When comparing paths you should lexicographically represent them as a sequence of characters, that is, of plot types.

Input

The first input line contains three integers n, m and k (1 ≤ n, m ≤ 50, n·m ≥ 2, 1 ≤ k ≤ 4). Then n lines contain the map. Each line has the length of exactly m characters and consists of lowercase Latin letters and characters S and T. It is guaranteed that the map contains exactly one character S and exactly one character T.

Pretest 12 is one of the maximal tests for this problem.

Output

If there is a path that satisfies the condition, print it as a sequence of letters — the plot types. Otherwise, print "-1" (without quotes). You shouldn't print the character S in the beginning and T in the end.

Note that this sequence may be empty. This case is present in pretests. You can just print nothing or print one "End of line"-character. Both will be accepted.

Examples

Input

5 3 2
Sba
ccc
aac
ccc
abT


Output

bcccc


Input

3 4 1
Sxyy
yxxx
yyyT


Output

xxxx


Input

1 3 3
TyS


Output

y


Input

1 4 1
SxyT


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const string filename = "TRACK";
int n, m, k;
char a[61][61];
bool mark[61];
int d1[61][61];
int d2[61][61];
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, 1, 0, -1};
int res_d = 1e9;
int sx, sy;
int tx, ty;
string res;
vector<pair<int, int> > tmp1;
vector<pair<int, int> > tmp2;
queue<pair<int, int> > CurrList;
void bfs() {
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) d1[i][j] = d2[i][j] = 1e9;
  int min_d = 1e9;
  string cur;
  CurrList.push(pair<int, int>(tx, ty));
  d2[tx][ty] = 0;
  while (!CurrList.empty()) {
    int x = CurrList.front().first;
    int y = CurrList.front().second;
    CurrList.pop();
    if (x == sx && y == sy) min_d = d2[x][y];
    for (int i = 0; i < 4; i++) {
      int xx = x + dx[i];
      int yy = y + dy[i];
      if (1 <= xx && xx <= n && 1 <= yy && yy <= m) {
        if (a[xx][yy] == 'S' || mark[a[xx][yy] - 'a' + 1]) {
          if (d2[x][y] + 1 < d2[xx][yy]) {
            d2[xx][yy] = d2[x][y] + 1;
            CurrList.push(pair<int, int>(xx, yy));
          }
        }
      }
    }
  }
  if (min_d == 1e9) return;
  tmp1.clear();
  tmp2.clear();
  tmp1.push_back(pair<int, int>(sx, sy));
  d1[sx][sy] = 0;
  for (int z = 1; z <= min_d; z++) {
    tmp2.clear();
    char tmp = 'z' + 1;
    for (int i = 0; i < tmp1.size(); i++) {
      int x = tmp1[i].first;
      int y = tmp1[i].second;
      if (x == tx && y == ty) break;
      for (int j = 0; j < 4; j++) {
        int xx = x + dx[j];
        int yy = y + dy[j];
        if (1 <= xx && xx <= n && 1 <= yy && yy <= m) {
          if (a[xx][yy] == 'T' || mark[a[xx][yy] - 'a' + 1]) {
            if ((d1[x][y] + 1 + d2[xx][yy] == min_d) && d1[xx][yy] == 1e9) {
              d1[xx][yy] = d1[x][y] + 1;
              if (a[xx][yy] < tmp) {
                tmp2.clear();
                tmp2.push_back(pair<int, int>(xx, yy));
                tmp = a[xx][yy];
              } else if (a[xx][yy] == tmp)
                tmp2.push_back(pair<int, int>(xx, yy));
            }
          }
        }
      }
    }
    cur += tmp;
    tmp1 = tmp2;
    tmp2.clear();
  }
  if (min_d < res_d || (min_d == res_d && cur < res)) {
    res_d = min_d;
    res = cur;
  }
}
void query(int cur, int used) {
  if (used <= k) bfs();
  if (used == k) return;
  for (int i = cur; i <= 26; i++) {
    mark[i] = true;
    query(i + 1, used + 1);
    mark[i] = false;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
      if (a[i][j] == 'S')
        sx = i, sy = j;
      else if (a[i][j] == 'T')
        tx = i, ty = j;
    }
  query(1, 0);
  if (res_d == 1e9)
    cout << -1;
  else {
    res.pop_back();
    cout << res;
  }
}