Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000000")
using namespace std;
const long long int INF = 10000;
int n, m, k, fx, fy, tx, ty;
pair<int, int> gone[4] = {make_pair(0, 1), make_pair(0, -1), make_pair(1, 0),
                          make_pair(-1, 0)};
bool correct(int x, int y) { return x >= 0 && y >= 0 && x < n && y < m; }
int ma[51][51];
int dest[51][51];
int check[51][51];
int used[51][51];
void ch_dfs(int x, int y) {
  if (dest[x][y] == INF) return;
  used[x][y]++;
  for (int i = 0; i < (4); i++) {
    int tox = x + gone[i].first;
    int toy = y + gone[i].second;
    if (correct(tox, toy) && !used[tox][toy] &&
        dest[tox][toy] == dest[x][y] - 1) {
      check[tox][toy] = check[x][y] + 1;
      ch_dfs(tox, toy);
    }
  }
}
bool comp(string &f, string &s) {
  if (s == "-") return true;
  return (f.length() < s.length() || (f.length() == s.length() && f < s));
}
string minans = "-";
pair<int, int> par[51][51];
void dfs(int x, int y) {
  string cur = "";
  vector<pair<int, pair<int, int> > > nex, f;
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) {
      par[i][j] = make_pair(-1, -1);
    }
  f.push_back(make_pair(ma[x][y], make_pair(x, y)));
  while (!f.empty()) {
    nex.clear();
    for (int i = 0; i < ((int)f.size()); i++) {
      if (i && f[i].first > f[i - 1].first) break;
      x = f[i].second.first;
      y = f[i].second.second;
      used[x][y]++;
      for (int i = 0; i < (4); i++) {
        int tox = x + gone[i].first;
        int toy = y + gone[i].second;
        if (correct(tox, toy) &&
            ((check[tox][toy] != -1 && check[tox][toy] == check[x][y] - 1) ||
             (tox == tx && toy == ty)) &&
            !used[tox][toy]) {
          par[tox][toy] = make_pair(x, y);
          used[tox][toy]++;
          nex.push_back(make_pair(ma[tox][toy], make_pair(tox, toy)));
        }
      }
    }
    sort(nex.begin(), nex.end());
    f = nex;
  }
  x = tx;
  y = ty;
  while (x != -1) {
    cur += (char)ma[x][y] + 'a';
    int nx = par[x][y].first;
    int ny = par[x][y].second;
    x = nx;
    y = ny;
  }
  reverse(cur.begin(), cur.end());
  if (comp(cur, minans)) {
    minans = cur;
  }
}
void bfs(int x, int y, vector<int> &cur) {
  map<int, int> good;
  for (int i = 0; i < ((int)cur.size()); i++) {
    good[cur[i]]++;
  }
  good[ma[x][y]]++;
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) {
      dest[i][j] = INF;
      check[i][j] = -1;
      used[i][j] = 0;
    }
  dest[x][y] = 0;
  dest[fx][fy] = 0;
  queue<pair<int, int> > q;
  q.push(make_pair(x, y));
  while (!q.empty()) {
    int xi = q.front().first;
    int yi = q.front().second;
    q.pop();
    for (int i = 0; i < (4); i++) {
      int tox = xi + gone[i].first;
      int toy = yi + gone[i].second;
      if (correct(tox, toy) &&
          (good.count(ma[tox][toy]) || (tox == tx && toy == ty)) &&
          dest[tox][toy] > dest[xi][yi] + 1) {
        dest[tox][toy] = dest[xi][yi] + 1;
        q.push(make_pair(tox, toy));
      }
    }
  }
  if (dest[tx][ty] == INF) return;
  check[tx][ty] = 0;
  ch_dfs(tx, ty);
  memset(used, 0, sizeof used);
  string curi = "";
  curi += ((char)'a' + ma[x][y]);
  dfs(x, y);
}
vector<pair<int, int> > fr;
void gen(int len, vector<int> &cur) {
  if (!len) {
    for (int i = 0; i < ((int)fr.size()); i++) {
      bfs(fr[i].first, fr[i].second, cur);
    }
    return;
  }
  int fr = (!(int)cur.size() ? 0 : cur.back() + 1);
  for (int i = fr; i <= 'z' - 'a'; i++) {
    cur.push_back(i);
    gen(len - 1, cur);
    cur.pop_back();
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) {
      char c;
      cin >> c;
      ma[i][j] = c - 'a';
      if (c == 'S') {
        fx = i;
        fy = j;
      }
      if (c == 'T') {
        tx = i;
        ty = j;
      }
    }
  for (int i = 0; i < (4); i++) {
    int tox = fx + gone[i].first;
    int toy = fy + gone[i].second;
    if (correct(tox, toy)) {
      fr.push_back(make_pair(tox, toy));
    }
  }
  vector<int> temp;
  gen(k - 1, temp);
  if (minans == "-")
    cout << -1;
  else {
    minans.resize((int)minans.size() - 1);
    cout << minans;
  }
}