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>
using namespace std;
template <typename T>
void cmax(T& a, T b) {
  a = max(a, b);
}
template <typename T>
void cmin(T& a, T b) {
  a = min(a, b);
}
void _BG(const char* s) {}
template <typename T, typename... TT>
void _BG(const char* s, T a, TT... b) {
  for (int c = 0; *s && (c || *s != ','); ++s) {
    cerr << *s;
    switch (*s) {
      case '(':
      case '[':
      case '{':
        ++c;
        break;
      case ')':
      case ']':
      case '}':
        --c;
        break;
    }
  }
  cerr << " = " << a;
  if (*s) {
    cerr << ", ";
    _BG(++s, b...);
  } else
    cerr << endl;
}
bool RD(void) { return true; }
bool RD(char& a) { return scanf(" %c", &a) == 1; }
bool RD(char* a) { return scanf("%s", a) == 1; }
bool RD(double& a) { return scanf("%lf", &a) == 1; }
bool RD(int& a) { return scanf("%d", &a) == 1; }
bool RD(long long& a) { return scanf("%lld", &a) == 1; }
template <typename T, typename... TT>
bool RD(T& a, TT&... b) {
  return RD(a) && RD(b...);
}
void PT(const char a) { putchar(a); }
void PT(const char* a) { fputs(a, stdout); }
void PT(char* a) { fputs(a, stdout); }
void PT(const double a) { printf("%.16f", a); }
void PT(const int a) { printf("%d", a); }
void PT(const long long a) { printf("%lld", a); }
template <char sep = ' ', char end = '\n'>
void PL(void) {
  if (end) PT(end);
}
template <char sep = ' ', char end = '\n', typename T, typename... TT>
void PL(const T a, const TT... b) {
  PT(a);
  if (sizeof...(b) && sep) PT(sep);
  PL<sep, end>(b...);
}
const int N = 55, inf = 5566, dr[] = {-1, 0, 0, 1}, dc[] = {0, -1, 1, 0};
int n, m, k;
vector<char> c;
char v[4];
bool w[26];
char g[N][N];
int d[N][N];
int vis[N][N], vc;
pair<int, int> s, t;
int al = inf - 1;
bool fl;
string ans;
bool ok(pair<int, int> x) {
  if (!(0 <= x.first && x.first < n && 0 <= x.second && x.second < m)) return 0;
  char ch = g[x.first][x.second];
  for (int i(0), _B(k); i < _B; ++i)
    if (ch == v[i]) return 1;
  return ch == 'S' || ch == 'T';
}
void bfs() {
  for (int i(0), _B(n); i < _B; ++i) fill_n(d[i], m, inf);
  queue<pair<int, int> > q;
  q.push(t);
  d[t.first][t.second] = 0;
  while (((int)q.size())) {
    pair<int, int> x = q.front();
    q.pop();
    for (int i(0), _B(4); i < _B; ++i) {
      pair<int, int> y(x.first + dr[i], x.second + dc[i]);
      if (ok(y) && d[y.first][y.second] == inf) {
        d[y.first][y.second] = d[x.first][x.second] + 1;
        q.push(y);
      }
    }
  }
  if (d[s.first][s.second] > al) return;
  al = d[s.first][s.second];
  string ca;
  vector<pair<int, int> > li[2];
  li[al & 1 ^ 1] = {s};
  for (int i((al)-1), _B(1); i >= _B; --i) {
    ++vc;
    ca += v[k - 1];
    for (auto x : li[i & 1]) {
      for (int j(0), _B(4); j < _B; ++j) {
        pair<int, int> y(x.first + dr[j], x.second + dc[j]);
        if (ok(y) && d[y.first][y.second] == i && vis[y.first][y.second] < vc) {
          vis[y.first][y.second] = vc;
          char ch = g[y.first][y.second];
          if (ch < ca.back()) {
            ca.back() = ch;
            li[i & 1 ^ 1].clear();
          }
          if (ch == ca.back()) li[i & 1 ^ 1].push_back(y);
        }
      }
    }
    li[i & 1].clear();
  }
  if (!fl || ((int)ans.size()) > ((int)ca.size()) || ans > ca) {
    ans = ca;
    fl = 1;
  }
}
void dfs(int i, int j) {
  if (i == k) {
    for (int u(0), _B(k); u < _B; ++u) bfs();
    return;
  }
  for (; ((int)c.size()) - j >= k - i; j++) {
    v[i] = c[j];
    dfs(i + 1, j + 1);
  }
}
int main() {
  RD(n, m, k);
  for (int i(0), _B(n); i < _B; ++i) {
    RD(g[i]);
    for (int j(0), _B(m); j < _B; ++j)
      if (g[i][j] == 'S')
        s = {i, j};
      else if (g[i][j] == 'T')
        t = {i, j};
      else
        w[g[i][j] - 'a'] = 1;
  }
  for (int i(0), _B(26); i < _B; ++i)
    if (w[i]) c.push_back(i + 'a');
  if (((int)c.size()) <= k) {
    k = ((int)c.size());
    for (int i(0), _B(k); i < _B; ++i) v[i] = c[i];
    bfs();
  } else {
    dfs(0, 0);
  }
  if (!fl)
    cout << "-1\n";
  else
    cout << ans << '\n';
}