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;
char g[55][55];
int nr, nc, k;
int tr, tc, sr, sc;
int mincost = -1;
char chosen[5];
bool seen[500];
char ans[3000];
int dr[] = {-1, 1, 0, 0};
int dc[] = {0, 0, 1, -1};
void check() {
  int d[55][55];
  memset((d), (-1), sizeof(d));
  d[tr][tc] = 0;
  queue<int> q;
  q.push(tr);
  q.push(tc);
  while (!q.empty()) {
    int r = q.front();
    q.pop();
    int c = q.front();
    q.pop();
    int dist = d[r][c] + 1;
    for (int i = 0, _e(4); i < _e; i++) {
      int rr = r + dr[i], cc = c + dc[i];
      if (rr < 0 || cc < 0 || rr >= nr || cc >= nc ||
          (g[rr][cc] != 'S' && !seen[g[rr][cc] - 'a']))
        continue;
      if (d[rr][cc] == -1 || d[rr][cc] > dist) {
        d[rr][cc] = dist;
        q.push(rr);
        q.push(cc);
      }
    }
  }
  if (d[sr][sc] != -1 && (mincost == -1 || mincost >= d[sr][sc])) {
    if (mincost == -1 || mincost > d[sr][sc]) {
      mincost = d[sr][sc];
      queue<int> q;
      int r = sr, c = sc;
      q.push(r);
      q.push(c);
      for (int j = 0, _e(mincost - 1); j < _e; j++) {
        int tot = q.size() / 2;
        char minC = 'z';
        queue<int> p;
        int taken[55][55] = {}, take = 1;
        for (int l = 0, _e(tot); l < _e; l++) {
          r = q.front();
          q.pop();
          c = q.front();
          q.pop();
          int look = mincost - j - 1;
          for (int i = 0, _e(4); i < _e; i++) {
            int rr = r + dr[i], cc = c + dc[i];
            if (rr < 0 || cc < 0 || rr >= nr || cc >= nc ||
                (!seen[g[rr][cc] - 'a']))
              continue;
            if (d[rr][cc] == look) {
              if (g[rr][cc] < minC) {
                while (p.size()) p.pop();
                ++take, minC = g[rr][cc], p.push(rr), p.push(cc),
                        taken[rr][cc] = take;
              } else if (g[rr][cc] == minC && taken[rr][cc] != take) {
                p.push(rr), p.push(cc), taken[rr][cc] = take;
              }
            }
          }
        }
        ans[j] = minC;
        q = p;
      }
    } else {
      queue<int> q;
      int r = sr, c = sc;
      q.push(r);
      q.push(c);
      char dans[3000];
      for (int j = 0, _e(mincost - 1); j < _e; j++) {
        int tot = q.size() / 2;
        char minC = 'z';
        queue<int> p;
        int taken[55][55] = {}, take = 0;
        for (int l = 0, _e(tot); l < _e; l++) {
          r = q.front();
          q.pop();
          c = q.front();
          q.pop();
          int look = mincost - j - 1;
          for (int i = 0, _e(4); i < _e; i++) {
            int rr = r + dr[i], cc = c + dc[i];
            if (rr < 0 || cc < 0 || rr >= nr || cc >= nc ||
                (!seen[g[rr][cc] - 'a']))
              continue;
            if (d[rr][cc] == look) {
              if (g[rr][cc] < minC) {
                while (p.size()) p.pop();
                ++take, minC = g[rr][cc], p.push(rr), p.push(cc),
                        taken[rr][cc] = take;
              } else if (g[rr][cc] == minC && taken[rr][cc] != take) {
                p.push(rr), p.push(cc), taken[rr][cc] = take;
              }
            }
          }
        }
        dans[j] = minC;
        q = p;
      }
      bool better = 0;
      for (int i = 0, _e(mincost - 1); i < _e; i++)
        if (ans[i] < dans[i])
          break;
        else if (ans[i] > dans[i])
          better = 1;
      if (better) {
        for (int i = 0, _e(mincost - 1); i < _e; i++) ans[i] = dans[i];
      }
    }
  }
}
void dfs(int p, int last) {
  if (p) check();
  if (p == k) return;
  for (int i(last), _e(25); i <= _e; i++)
    if (!seen[i]) {
      seen[i] = 1;
      chosen[p] = i + 'a';
      dfs(p + 1, i + 1);
      seen[i] = 0;
    }
}
int main() {
  cin >> nr >> nc >> k;
  for (int i = 0, _e(nr); i < _e; i++) cin >> g[i];
  for (int i = 0, _e(nr); i < _e; i++)
    for (int j = 0, _e(nc); j < _e; j++)
      if (g[i][j] == 'T')
        tr = i, tc = j;
      else if (g[i][j] == 'S')
        sr = i, sc = j;
  dfs(0, 0);
  if (mincost == -1) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 0, _e(mincost - 1); i < _e; i++) cout << ans[i];
  cout << endl;
  return 0;
}
