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 int MAXN = 55;
struct mpair {
  string s;
  int x, y, m, c;
  mpair(string S = "", int X = 0, int Y = 0, int M = 0, int C = 0)
      : s(S), x(X), y(Y), m(M), c(C) {}
};
bool operator<(const mpair &a, const mpair &b) {
  return a.s.size() > b.s.size() || a.s.size() == b.s.size() && a.s > b.s;
}
const int xi[4] = {-1, 0, 1, 0}, yi[4] = {0, -1, 0, 1};
set<long long> S;
priority_queue<mpair> H;
int n, m, k;
int xs, ys;
char a[MAXN][MAXN];
int o[MAXN][MAXN];
string ans;
int add(int i, int j, string s, int m, int c) {
  if (c > k) return 0;
  long long x = (((((long long)m) << 6) + i) << 6) + j;
  if (S.find(x) != S.end()) return 0;
  if (o[i][j] && ((o[i][j] & m) == o[i][j])) return 0;
  o[i][j] = m;
  S.insert(x);
  H.push(mpair(s, i, j, m, c));
  return 0;
}
int main() {
  scanf("%d %d %d\n", &n, &m, &k);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%c", &a[i][j]);
      if (a[i][j] == 'S') {
        xs = i;
        ys = j;
      }
    }
    scanf("\n");
  }
  add(xs, ys, "", 0, 0);
  while (H.size()) {
    mpair p = H.top();
    H.pop();
    string s = p.s;
    int x = p.x, y = p.y, mask = p.m, count = p.c;
    for (int i = 0; i < 4; i++) {
      int xx = x + xi[i], yy = y + yi[i];
      if (xx < 1 || n < xx || yy < 1 || m < yy || a[xx][yy] == 'S') continue;
      if (a[xx][yy] == 'T') {
        cout << s << endl;
        return 0;
      }
      int j = a[xx][yy] - 'a';
      add(xx, yy, s + a[xx][yy], mask | (1 << j),
          count + (((1 << j) & mask) == 0));
    }
  }
  puts("-1");
  return 0;
}
