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;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, -1, 1};
int n, m, k, sx, sy, fx, fy;
string s[60], a[60][60], res;
vector<int> qx, qy;
bool ok[26];
bool valid(int x, int y) {
  if (x >= 0 && y >= 0 && x < n && y < m) return 1;
  return 0;
}
void bfs() {
  for (int i = 0; i < qx.size(); i++) {
    int x, y;
    for (int j = 0; j < 4; j++) {
      x = qx[i] + dx[j];
      y = qy[i] + dy[j];
      if (valid(x, y) &&
          (s[x][y] == 'T' || s[x][y] == 'S' || ok[s[x][y] - 'a'])) {
        if (a[x][y].size() == 0) {
          qx.push_back(x);
          qy.push_back(y);
          a[x][y] = a[qx[i]][qy[i]] + s[x][y];
        } else if ((a[x][y].size() > a[qx[i]][qy[i]].size() + 1) ||
                   (a[x][y].size() == a[qx[i]][qy[i]].size() + 1 &&
                    a[x][y] > a[qx[i]][qy[i]] + s[x][y])) {
          a[x][y] = a[qx[i]][qy[i]] + s[x][y];
        }
      }
    }
  }
}
void f(int x, int y) {
  if (x == 0) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) a[i][j].clear();
    qx.clear();
    qy.clear();
    qx.push_back(sx);
    qy.push_back(sy);
    bfs();
    if (a[fx][fy].size()) {
      if (!res.size() || res.size() > a[fx][fy].size() ||
          (a[fx][fy].size() == res.size() && res > a[fx][fy])) {
        res = a[fx][fy];
      }
    }
  } else {
    for (int i = y; i < 26; i++) {
      ok[i] = 1;
      f(x - 1, i + 1);
      ok[i] = 0;
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      if (s[i][j] == 'S') {
        sx = i;
        sy = j;
      }
      if (s[i][j] == 'T') {
        fx = i;
        fy = j;
      }
    }
  f(k, 0);
  if (res.size()) {
    for (int i = 0; i < res.size() - 1; i++) cout << res[i];
    cout << endl;
  } else
    cout << -1 << endl;
  return 0;
}
