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
#include <bits/stdc++.h>
using namespace std;
struct point {
  char h;
  string res;
  bool poss;
};
point a[50][50];
int n, m, k;
string res;
vector<int> qx, qy;
int step[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
void bfs() {
  for (int i = 0; i < qx.size(); i++) {
    int x, y;
    for (int j = 0; j < 4; j++) {
      x = qx[i] + step[j][0];
      y = qy[i] + step[j][1];
      if (x > -1 && y > -1 && x < n && y < m && a[x][y].poss) {
        if (a[x][y].res.size() == 0) {
          qx.push_back(x);
          qy.push_back(y);
          a[x][y].res = a[qx[i]][qy[i]].res + a[x][y].h;
        } else {
          if (a[x][y].res.size() == a[qx[i]][qy[i]].res.size() + 1 &&
              a[x][y].res > a[qx[i]][qy[i]].res + a[x][y].h) {
            a[x][y].res = a[qx[i]][qy[i]].res + a[x][y].h;
          }
        }
      }
    }
  }
}
vector<char> temp, meeted;
char now[6];
int startx, starty, fx, fy;
void pere(int deep, int start) {
  for (int i = start; i < meeted.size(); i++) {
    now[deep + 2] = meeted[i];
    if (deep == k - 1) {
      qx.clear();
      qy.clear();
      qx.push_back(startx);
      qy.push_back(starty);
      for (int ii = 0; ii < n; ii++)
        for (int jj = 0; jj < m; jj++) {
          a[ii][jj].res.clear();
          a[ii][jj].poss = false;
          for (int j = 0; j < k + 2; j++)
            if (a[ii][jj].h == now[j]) a[ii][jj].poss = true;
        }
      a[startx][starty].res = "S";
      bfs();
      if (a[fx][fy].res.size()) {
        if (!res.size() || res.size() > a[fx][fy].res.size() ||
            (a[fx][fy].res.size() == res.size() && res > a[fx][fy].res))
          res = a[fx][fy].res;
      }
    } else {
      pere(deep + 1, i + 1);
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> a[i][j].h;
      if (a[i][j].h == 'S') {
        startx = i;
        starty = j;
      }
      if (a[i][j].h == 'T') {
        fx = i;
        fy = j;
      }
      if (a[i][j].h >= 'a') temp.push_back(a[i][j].h);
    }
  }
  now[0] = 'S';
  now[1] = 'T';
  sort(temp.begin(), temp.end());
  for (int i = 0; i < temp.size(); i++) {
    if (!i || temp[i] != temp[i - 1]) meeted.push_back(temp[i]);
  }
  k = min(k, int(meeted.size()));
  if (meeted.size() == 0) return 0;
  pere(0, 0);
  if (res.size() == 0)
    cout << -1;
  else {
    for (int i = 1; i < res.size() - 1; i++) cout << res[i];
  }
}
