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;
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename C>
ostream& operator<<(ostream& os, const vector<C>& v) {
  for (__typeof((v).begin()) __it = (v).begin(); __it != (v).end(); __it++)
    os << *(__it) << ' ';
  return os;
}
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;
const int maxn = 52;
char field[maxn][maxn];
int dir[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
vector<pair<int, int> > q[maxn * maxn * 4];
bool visit[maxn][maxn];
pair<int, int> from[maxn][maxn];
int n, m, k;
pair<int, int> start, finish;
int letter_can[26];
int letter_canf[26];
inline bool check(const pair<int, int>& p) {
  if (p.first < 0 || p.first >= n) return false;
  if (p.second < 0 || p.second >= m) return false;
  return true;
}
char seq[5];
int can_start, can_finish;
bool finalresset;
string finalres;
void dfs(int letter, int curk) {
  if (curk < k) {
    for (int i = letter; i <= 'z'; i++) {
      seq[curk] = i;
      letter_canf[i - 'a']++;
      if (letter_can[i - 'a'] / 10) can_start++;
      if (letter_can[i - 'a'] % 10) can_finish++;
      dfs(i, curk + 1);
      if (letter_can[i - 'a'] / 10) can_start--;
      if (letter_can[i - 'a'] % 10) can_finish--;
      letter_canf[i - 'a']--;
    }
    return;
  }
  if (can_start * can_finish == 0) return;
  ;
  int bc = 0;
  int fr = 1;
  memset(visit, 0, sizeof(visit));
  q[0].clear();
  visit[start.first][start.second] = true;
  q[0].push_back(start);
  from[finish.first][finish.second] = pair<int, int>(-1, -1);
  while (bc < fr) {
    ;
    ;
    vector<pair<int, int> >& c = q[bc++];
    vector<pair<int, int> > qnext[4];
    for (int i = 0; i < (int((c).size())); i++) {
      ;
      for (int j = 0; j < 4; j++) {
        ;
        pair<int, int> next = c[i];
        next.first += dir[j][0];
        next.second += dir[j][1];
        if (!check(next)) continue;
        char nextc = field[next.first][next.second];
        ;
        if ((letter_canf[nextc - 'a'] || nextc == 'T') &&
            !visit[next.first][next.second]) {
          visit[next.first][next.second] = true;
          from[next.first][next.second] = c[i];
          if (nextc == 'T') continue;
          int w;
          for (w = 0; seq[w] != nextc; w++)
            ;
          qnext[w].push_back(next);
        }
      }
    }
    for (int i = 0; i < 4; i++)
      if (qnext[i].size()) q[fr++] = qnext[i];
  };
  if (from[finish.first][finish.second] == pair<int, int>(-1, -1)) return;
  string res;
  pair<int, int> cur = from[finish.first][finish.second];
  while (cur != start) {
    res += field[cur.first][cur.second];
    cur = from[cur.first][cur.second];
  }
  reverse(res.begin(), res.end());
  if (!finalresset || (int((res).size())) < (int((finalres).size())) ||
      ((int((res).size())) == (int((finalres).size())) && res < finalres)) {
    finalres = res;
    finalresset = true;
  }
  return;
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < n; i++) {
    scanf("%s", field[i]);
    for (int j = 0; j < m; j++)
      if (field[i][j] == 'S')
        start = pair<int, int>(i, j);
      else if (field[i][j] == 'T')
        finish = pair<int, int>(i, j);
  }
  for (int i = 0; i < 4; i++) {
    pair<int, int> next = start;
    next.first += dir[i][0];
    next.second += dir[i][1];
    if (next == finish) {
      cout << endl;
      return 0;
    }
    if (check(next) && next != finish) {
      letter_can[field[next.first][next.second] - 'a'] += 10;
    }
    next = finish;
    next.first += dir[i][0];
    next.second += dir[i][1];
    if (check(next) && next != start) {
      letter_can[field[next.first][next.second] - 'a'] += 1;
    }
  }
  dfs('a', 0);
  if (finalresset)
    cout << finalres << endl;
  else
    cout << -1 << endl;
  return 0;
}
