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 n, m, k, sx, sy, ex, ey;
bool valid[127];
char s[55][55];
int q[55 * 55 * 2], op, cl;
int dis[55][55];
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
bool good[55][55];
string out;
void bfs2() {
  string res = "";
  memset(good, false, sizeof(good));
  op = cl = 0;
  q[op++] = ex;
  q[op++] = ey;
  good[ex][ey] = true;
  char bes = 'S', prev = 'z';
  int d = dis[ex][ey];
  while (op != cl) {
    int x = q[cl++], y = q[cl++];
    if (dis[x][y] != d) {
      --d;
      if (bes != 'S') res += bes;
      bes = prev;
      prev = 'z';
    }
    if (x == sx && y == sy) break;
    if (s[x][y] != bes) continue;
    for (int i = 0; i < 4; i++) {
      int xx = x - dx[i], yy = y - dy[i];
      if (xx >= 0 && xx < n && yy >= 0 && yy < m &&
          dis[xx][yy] == dis[x][y] - 1 && !good[xx][yy]) {
        good[xx][yy] = true;
        q[op++] = xx;
        q[op++] = yy;
        if (s[xx][yy] < prev) prev = s[xx][yy];
      }
    }
  }
  if (out == "-1" || out.length() > res.length() ||
      out.length() == res.length() && out > res)
    out = res;
}
void bfs() {
  op = cl = 0;
  q[op++] = sx;
  q[op++] = sy;
  memset(dis, -1, sizeof(dis));
  dis[sx][sy] = 0;
  while (op != cl) {
    int x = q[cl++], y = q[cl++];
    if (x == ex && y == ey) {
      bfs2();
      return;
    }
    for (int i = 0; i < 4; i++) {
      int xx = x + dx[i], yy = y + dy[i];
      if (xx >= 0 && xx < n && yy >= 0 && yy < m) {
        if (valid[s[xx][yy]] && dis[xx][yy] == -1) {
          q[op++] = xx;
          q[op++] = yy;
          dis[xx][yy] = dis[x][y] + 1;
        }
      }
    }
  }
}
void dfs(int i, char s) {
  if (i == k) {
    bfs();
    return;
  }
  while (s <= 'z') {
    valid[s] = true;
    dfs(i + 1, s + 1);
    valid[s] = false;
    ++s;
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%s", s[i]);
    for (int j = 0; j < m; j++)
      if (s[i][j] == 'T') {
        sx = i;
        sy = j;
      } else if (s[i][j] == 'S') {
        ex = i;
        ey = j;
      }
  }
  valid['S'] = true;
  out = "-1";
  dfs(0, 'a');
  puts(out.c_str());
  return 0;
}
