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:134217728")
using namespace std;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
int n;
int m;
int matr[50][50];
bool cmp(string a, string b) {
  if (((int)(a).size()) != ((int)(b).size()))
    return ((int)(a).size()) < ((int)(b).size());
  return a < b;
}
int x1, y95874, x2, y2;
int d[50][50];
queue<pair<int, int> > q;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int mask;
int good(int x, int y) {
  return x >= 0 && x < n && y >= 0 && y < m &&
         (make_pair(x1, y95874) == make_pair(x, y) ||
          make_pair(x2, y2) == make_pair(x, y) ||
          ((1 << matr[x][y]) & mask) != 0);
}
void parse(pair<int, int> o) {
  int x = o.first;
  int y = o.second;
  int dist = d[x][y];
  for (int i = 0; i < (4); i++) {
    int nx = x + dx[i];
    int ny = y + dy[i];
    if (good(nx, ny) && d[nx][ny] == -1) {
      d[nx][ny] = dist + 1;
      q.push(make_pair(nx, ny));
    }
  }
}
string getans(int mask) {
  ::mask = mask;
  memset(d, -1, sizeof(d));
  q.push(make_pair(x2, y2));
  d[x2][y2] = 0;
  while (!q.empty()) {
    parse(q.front());
    q.pop();
  }
  if (d[x1][y95874] == -1) return "-";
  int cx = x1, cy = y95874;
  string ans = "";
  vector<pair<int, int> > v;
  v.push_back(make_pair(x1, y95874));
  vector<pair<int, int> > tmp;
  tmp.reserve(1000);
  while (1) {
    int dist = d[v[0].first][v[0].second];
    if (dist == 1) break;
    char bc = 'z' + 1;
    for (int o = 0; o < (((int)(v).size())); o++)
      for (int i = 0; i < (4); i++) {
        int nx = v[o].first + dx[i];
        int ny = v[o].second + dy[i];
        if (good(nx, ny) && d[nx][ny] == dist - 1) {
          if (matr[nx][ny] + 'a' < bc) {
            bc = matr[nx][ny] + 'a';
          }
        }
      }
    tmp.clear();
    for (int o = 0; o < (((int)(v).size())); o++)
      for (int i = 0; i < (4); i++) {
        int nx = v[o].first + dx[i];
        int ny = v[o].second + dy[i];
        if (good(nx, ny) && d[nx][ny] == dist - 1) {
          if (matr[nx][ny] + 'a' == bc) {
            tmp.push_back(make_pair(nx, ny));
          }
        }
      }
    sort((tmp).begin(), (tmp).end());
    tmp.resize(unique((tmp).begin(), (tmp).end()) - tmp.begin());
    ans += bc;
    v = tmp;
  }
  return ans;
}
int main() {
  int k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < (n); i++) {
    char s[1000];
    scanf("%s", s);
    for (int j = 0; j < (m); j++) {
      matr[i][j] = s[j] - 'a';
      if (s[j] == 'S') {
        x1 = i;
        y95874 = j;
      }
      if (s[j] == 'T') {
        x2 = i;
        y2 = j;
      }
    }
  }
  vector<int> v;
  v.reserve(20000);
  for (int i = 0; i < ((1 << 26)); i++) {
    if (__builtin_popcount(i) <= k) v.push_back(i);
  }
  string res = "-";
  for (int i = 0; i < (((int)(v).size())); i++) {
    string ans = getans(v[i]);
    if (ans != "-")
      if (res == "-" || cmp(ans, res)) res = ans;
  }
  if (res == "-")
    cout << -1 << endl;
  else
    cout << res << endl;
  return 0;
}