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;
string res = "-1", cur;
const int dr[4] = {-1, 0, 1, 0};
const int dc[4] = {0, 1, 0, -1};
int R, C, k, sr, sc, tr, tc, q[100000], f, l, r, c, nr, nc, d;
vector<int> p[50][50];
char s[50][51];
int cost[50][50];
bool in[27];
vector<int> push;
void build() {
  q[f = l = 0] = tr;
  q[++l] = tc;
  cur = "";
  while (f <= l) {
    char mn = 28;
    push.clear();
    while (f <= l) {
      r = q[f];
      ++f;
      c = q[f];
      ++f;
      for (int i = 0; i < p[r][c].size(); ++i) {
        nr = r + dr[p[r][c][i]];
        nc = c + dc[p[r][c][i]];
        if (cost[nr][nc] == -1) continue;
        cost[nr][nc] = -1;
        if (s[nr][nc] < mn) {
          mn = s[nr][nc];
          push.clear();
        }
        if (s[nr][nc] == mn) {
          push.push_back(nr);
          push.push_back(nc);
        }
      }
    }
    if (mn == 26) break;
    cur += mn + 'a';
    for (int i = 0; i < push.size(); ++i) q[++l] = push[i];
  }
  if (res == "-1" || res.size() > cur.size() ||
      res.size() == cur.size() && res > cur)
    res = cur;
}
void BFS() {
  for (int i = 0; i < R; ++i)
    for (int j = 0; j < C; ++j) {
      p[i][j].clear();
      cost[i][j] = -1;
    }
  q[f = l = 0] = sr;
  q[++l] = sc;
  cost[sr][sc] = 0;
  while (f <= l) {
    r = q[f];
    ++f;
    c = q[f];
    ++f;
    if (r == tr && c == tc) {
      build();
      return;
    }
    for (int i = 0; i < 4; ++i) {
      nr = r + dr[i];
      nc = c + dc[i];
      if (nr < 0 || nr == R || nc < 0 || nc == C || !in[s[nr][nc]]) continue;
      if (cost[nr][nc] == -1) {
        cost[nr][nc] = cost[r][c] + 1;
        p[nr][nc].push_back((i + 2) & 3);
        q[++l] = nr;
        q[++l] = nc;
      } else if (cost[nr][nc] == 1 + cost[r][c])
        p[nr][nc].push_back((i + 2) & 3);
    }
  }
}
void calc(int i, int rem) {
  if (!rem) {
    BFS();
    return;
  }
  if (i == 26) return;
  in[i] = true;
  calc(i + 1, rem - 1);
  in[i] = false;
  calc(i + 1, rem);
}
int main() {
  scanf("%d%d%d", &R, &C, &k);
  for (int i = 0; i < R; ++i) {
    scanf("%s", s[i]);
    for (int j = 0; j < C; ++j)
      if (s[i][j] == 'S') {
        tr = i;
        tc = j;
        s[i][j] = 26;
      } else if (s[i][j] == 'T') {
        sr = i;
        sc = j;
        s[i][j] = 26;
      } else
        s[i][j] -= 'a';
  }
  in[26] = true;
  calc(0, k);
  puts(res.c_str());
  return 0;
}
