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;
const double eps = 1e-10;
const int oo = ~0u >> 2, mo = (int)1e9 + 7;
const int mn = 60, mc = 26, ml = 14, mm = 1 << ml;
const int fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
int a[mn][mn];
int Cnt[mm];
int n, m, k;
pair<int, int> st, ed;
int Encode(pair<int, int> t) { return t.first * m + t.second; }
pair<int, int> Decode(int t) { return make_pair(t / m, t % m); }
int dis(pair<int, int> x, pair<int, int> y) {
  return abs(x.first - y.first) + abs(x.second - y.second);
}
int cntb(int x) {
  int ret = 0;
  while (x) ret += Cnt[x & (mm - 1)], x >>= ml;
  return ret;
}
struct Node {
  string path;
  int col, posi;
  Node(string _path, int _col, int _posi) {
    path = _path, col = _col, posi = _posi;
  }
  long long Encode() { return (long long)col * 10000 + posi; }
  bool operator<(const Node &p) const {
    pair<int, int> p1 = Decode(posi), p2 = Decode(p.posi);
    int d1 = (int)path.length() + dis(p1, ed);
    int d2 = (int)p.path.length() + dis(p2, ed);
    if (d1 != d2) return d1 > d2;
    return path > p.path;
  }
};
priority_queue<Node> Q;
set<long long> dup;
bool okay(0);
int main() {
  for (int i = (1); i <= (mm - 1); ++i) Cnt[i] = __builtin_popcount(i);
  cin >> n >> m >> k;
  for (int i = (0); i <= (n - 1); ++i)
    for (int j = (0); j <= (m - 1); ++j) {
      char ch;
      do {
        scanf("%c", &ch);
      } while ((ch < 'a' || ch > 'z') && ch != 'S' && ch != 'T');
      if (ch == 'S')
        st = make_pair(i, j), a[i][j] = mc;
      else if (ch == 'T')
        ed = make_pair(i, j), a[i][j] = mc + 1;
      else
        a[i][j] = ch - 'a';
    }
  Q.push(Node("", 0, Encode(st)));
  while (!Q.empty() && !okay) {
    Node Now = Q.top();
    Q.pop();
    pair<int, int> P = Decode(Now.posi);
    if (a[P.first][P.second] > mc) {
      cout << Now.path << endl;
      okay = 1;
      break;
    }
    if (dup.find(Now.Encode()) != dup.end()) continue;
    dup.insert(Now.Encode());
    for (int i = (0); i <= (3); ++i) {
      pair<int, int> p = P;
      p.first += fx[i][0], p.second += fx[i][1];
      if (p.first < 0 || p.first >= n || p.second < 0 || p.second >= m)
        continue;
      if (a[p.first][p.second] == mc) continue;
      string np = Now.path;
      int col = Now.col;
      if (a[p.first][p.second] < mc)
        np += char('a' + a[p.first][p.second]),
            col |= 1 << a[p.first][p.second];
      Node tmp = Node(np, col, Encode(p));
      if (cntb(tmp.col) > k) continue;
      Q.push(tmp);
    }
  }
  if (!okay) cout << -1 << endl;
}
